Question 1

Q1. A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Accepted Answer

The magnitude of the magnetic field at the centre of a circular coil with N turns, radius R, and carrying a current I is given by the formula B = (μ₀ * N * I) / (2 * R). By substituting the given values, we can calculate the magnetic field strength.

Given:
Number of turns, N = 100
Radius of the coil, R = 8.0 cm = 0.08 m
Current in the coil, I = 0.40 A
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m A⁻¹

Calculation:
B = (4π × 10⁻⁷ T m A⁻¹ × 100 × 0.40 A) / (2 × 0.08 m)
B = (4π × 10⁻⁷ × 40) / 0.16 T
B = (160π × 10⁻⁷) / 0.16 T
B = 1000π × 10⁻⁷ T
B = π × 10⁻⁴ T
B ≈ 3.14 × 10⁻⁴ T

Thus, the magnitude of the magnetic field at the centre of the coil is 3.14 × 10⁻⁴ T.

Question 2

Q2. A long straight wire carries a current of 35 A. What is the magnitude of the magnetic field B at a point 20 cm from the wire?

Accepted Answer

The magnitude of the magnetic field at a perpendicular distance 'r' from a long, straight wire carrying a current 'I' is given by Ampere's Law, B = (μ₀ * I) / (2 * π * r).

Given:
Current in the wire, I = 35 A
Distance from the wire, r = 20 cm = 0.20 m
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m A⁻¹

Calculation:
B = (4π × 10⁻⁷ T m A⁻¹ × 35 A) / (2 * π * 0.20 m)
B = (2 × 10⁻⁷ × 35) / 0.20 T
B = (70 × 10⁻⁷) / 0.20 T
B = 350 × 10⁻⁷ T
B = 3.5 × 10⁻⁵ T

Therefore, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 × 10⁻⁵ T.

Question 3

Q3. A long straight wire in the horizontal plane carries a current of 50 A in the north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

Accepted Answer

This problem requires calculating the magnitude of the magnetic field using the formula for a long straight wire and determining its direction using the Right-Hand Thumb Rule.

**1. Magnitude of the Magnetic Field (B)**

Given:
Current (I) = 50 A
Distance (r) = 2.5 m
Permeability of free space (μ₀) = 4π × 10⁻⁷ T m A⁻¹

Formula: B = (μ₀ * I) / (2 * π * r)

Calculation:
B = (4π × 10⁻⁷ T m A⁻¹ × 50 A) / (2 * π * 2.5 m)
B = (2 × 10⁻⁷ × 50) / 2.5 T
B = (100 × 10⁻⁷) / 2.5 T
B = 40 × 10⁻⁷ T
B = 4 × 10⁻⁶ T

**2. Direction of the Magnetic Field (B)**

We use the Right-Hand Thumb Rule:
- Point the thumb of your right hand in the direction of the current, which is from North to South.
- The point of observation is 2.5 m East of the wire.
- Curl your fingers around the wire. At the point to the East, your fingers will point vertically upwards.

Therefore, the direction of the magnetic field is vertically upwards.

**Final Answer:** The magnitude of the magnetic field is 4 × 10⁻⁶ T, and its direction is vertically upwards.

Question 4

Q4. A horizontal overhead power line carries a current of 90 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below t

Accepted Answer

This problem involves finding the magnitude and direction of the magnetic field produced by a long, straight current-carrying wire.

**1. Magnitude of the Magnetic Field (B)**

Given:
Current (I) = 90 A
Distance (r) = 1.5 m
Permeability of free space (μ₀) = 4π × 10⁻⁷ T m A⁻¹

Formula: B = (μ₀ * I) / (2 * π * r)

Calculation:
B = (4π × 10⁻⁷ T m A⁻¹ × 90 A) / (2 * π * 1.5 m)
B = (2 × 10⁻⁷ × 90) / 1.5 T
B = (180 × 10⁻⁷) / 1.5 T
B = 120 × 10⁻⁷ T
B = 1.2 × 10⁻⁵ T

**2. Direction of the Magnetic Field (B)**

We use the Right-Hand Thumb Rule:
- The current flows from East to West.
- Point the thumb of your right hand in the direction of the current (towards the West).
- The point of observation is 1.5 m directly below the wire.
- Curl your fingers. At the point below the wire, your fingers will point towards the South.

Therefore, the direction of the magnetic field is towards the South.

**Final Answer:** The magnitude of the magnetic field is 1.2 × 10⁻⁵ T, and its direction is towards the South.

Question 5

Q5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Accepted Answer

The magnetic force (F) on a straight wire of length L carrying a current I, placed in a uniform magnetic field B, is given by the formula F = I * L * B * sin(θ), where θ is the angle between the direction of the current and the magnetic field.

The question asks for the force per unit length (f), which is F/L.
So, f = F/L = I * B * sin(θ).

Given:
Current (I) = 8 A
Magnetic field (B) = 0.15 T
Angle (θ) = 30°

Calculation:
f = 8 A × 0.15 T × sin(30°)
We know that sin(30°) = 0.5
f = 8 × 0.15 × 0.5 N/m
f = 8 × 0.075 N/m
f = 0.6 N/m

Thus, the magnitude of the magnetic force per unit length on the wire is 0.6 N/m.

Question 6

Q6. A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an

Accepted Answer

To prevent the coil from turning, a counter torque equal in magnitude and opposite in direction to the magnetic torque acting on the coil must be applied. The magnetic torque on a current-carrying coil in a uniform magnetic field is given by the formula τ = NIAB sinθ, where N is the number of turns, I is the current, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the normal to the coil and the magnetic field.

Question 7

Q7. A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of wire are wound. If the current in the wire is 11 A, what is the mag

Accepted Answer

The magnetic field produced by a toroid varies depending on the region. For a toroid, the magnetic field is non-zero only within the core of the toroid. Outside the toroid and in the empty space enclosed by the toroid (the hole), the magnetic field is zero. The magnitude of the magnetic field inside the core is determined using Ampere's circuital law.

Question 8

Q8. A chamber has a uniform magnetic field of 6.5 G (1 G = 10⁻⁴ T) and an electron enters into the chamber with a speed of 4.8 × 10⁶ m s⁻¹ normal to the field.
Explain why the path of

Accepted Answer

When a charged particle enters a uniform magnetic field perpendicular to its velocity, the magnetic force acts as a centripetal force, causing the particle to move in a circular path. The magnetic force is always perpendicular to both the velocity of the particle and the magnetic field, thus it changes the direction of the velocity without changing its magnitude (speed).

Question 9

Q9. In Question 4.8, obtain the frequency of revolution of the electron in its circular orbit. Does the frequency depend on the speed of the electron?

Accepted Answer

The frequency of revolution of a charged particle in a uniform magnetic field is also known as the cyclotron frequency. It can be derived from the relationship between the radius, speed, and time period of circular motion. An interesting characteristic of this frequency is its independence from the particle's speed and the radius of its orbit, as long as the particle is non-relativistic.

Question 10

Q10. A circular coil of 16 turns and radius 10 cm is placed in a uniform magnetic field of 0.05 T normal to the coil. If the current in the coil is 10 A, calculate
(a) the total torque

Accepted Answer

This question explores multiple aspects of a current-carrying coil in a magnetic field: the torque experienced by the coil, the net force on the coil, and the force on individual charge carriers (electrons) within the coil. The key is to understand the conditions under which each of these forces/torques arises.

Question 11

Q11. A magnetic field of 100 G (1 G = 10⁻⁴ T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10⁻³ m². The maximum current carry

Accepted Answer

To achieve a uniform magnetic field of 100 G (10⁻² T) using a solenoid with a non-ferromagnetic core, we can use the formula for the magnetic field inside a solenoid: B = μ₀nI.

Given:
Required magnetic field, B = 100 G = 100 × 10⁻⁴ T = 10⁻² T
Maximum current, I_max = 15 A
Maximum turns per unit length, n_max = 1000 turns/m
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m/A

From B = μ₀nI, we need the product nI to be:
nI = B / μ₀ = (10⁻² T) / (4π × 10⁻⁷ T m/A) ≈ 7957.75 A/m

We need to choose values for n and I such that their product is approximately 7958 A/m, while respecting the maximum limits.

Let's choose a current I that is less than or equal to I_max = 15 A. For example, let's choose I = 10 A.
Then, the required turns per unit length would be:
n = (nI) / I = 7957.75 A/m / 10 A = 795.775 turns/m

This value of n (approximately 796 turns/m) is well within the maximum allowed turns per unit length (n_max = 1000 turns/m).

Alternatively, if we choose to use the maximum allowed turns per unit length, n = 1000 turns/m, then the required current would be:
I = (nI) / n = 7957.75 A/m / 1000 turns/m = 7.95775 A

This value of I (approximately 7.96 A) is well within the maximum allowed current (I_max = 15 A).

**Suggested Design Particulars:**
1.  **Option 1 (Optimizing for current):**
    *   Current (I): 10 A
    *   Number of turns per unit length (n): Approximately 796 turns/m
    *   Total number of turns for a 10 cm (0.1 m) length: 796 turns/m × 0.1 m ≈ 80 turns

2.  **Option 2 (Optimizing for turns per unit length):**
    *   Current (I): Approximately 7.96 A
    *   Number of turns per unit length (n): 1000 turns/m
    *   Total number of turns for a 10 cm (0.1 m) length: 1000 turns/m × 0.1 m = 100 turns

Both options are feasible. Option 2 uses the maximum winding density, which might be easier to achieve if the wire gauge allows it, and requires less current. Option 1 uses a slightly higher current but a lower winding density.

For practical purposes, a solenoid of length slightly greater than 10 cm (e.g., 15-20 cm) with the required number of turns per unit length would ensure a uniform field over the desired 10 cm region near its center. The area of cross-section (10⁻³ m²) is relevant for the overall size but doesn't directly affect the B-field calculation for an ideal solenoid.

Question 12

Q12. For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by B = (μ₀NIR²)/

Accepted Answer

The magnetic field on the axis of a circular coil of radius R, N turns, and current I at a distance x from its center is given by:
B = (μ₀NIR²) / (2(x² + R²)^(3/2))

**(a) Field at the centre of the coil (x = 0)**
To find the magnetic field at the center of the coil, we substitute x = 0 into the given formula:
B_center = (μ₀NIR²) / (2(0² + R²)^(3/2))
B_center = (μ₀NIR²) / (2(R²)^(3/2))
B_center = (μ₀NIR²) / (2R³)
B_center = (μ₀NI) / (2R)
This is the familiar result for the magnetic field at the center of a circular coil.

**(b) Helmholtz Coils**
Consider two parallel co-axial circular coils, each of radius R and N turns, carrying current I in the same direction, separated by a distance R. Let the origin (x=0) be at the center of the first coil. Then the first coil is at x = 0 and the second coil is at x = R.

Let's consider a point P on the axis at a distance x' from the center of the first coil. The distance of P from the center of the first coil is x', and from the center of the second coil is (R - x').

The magnetic field at point P due to the first coil (B₁) is:
B₁ = (μ₀NIR²) / (2(x'² + R²)^(3/2))

The magnetic field at point P due to the second coil (B₂) is:
B₂ = (μ₀NIR²) / (2((R - x')² + R²)^(3/2))

Since the currents are in the same direction, the fields add up. The total magnetic field B_total at point P is:
B_total = B₁ + B₂ = (μ₀NIR²/2) * [ 1/(x'² + R²)^(3/2) + 1/((R - x')² + R²)^(3/2) ]

We are interested in the magnetic field around the mid-point between the coils. The mid-point is at x' = R/2. Let's consider a small displacement 'y' from the mid-point, so x' = R/2 + y.

To show that the field is uniform over a small distance around the mid-point, we need to show that the first derivative of B_total with respect to x' (or y) is zero at the mid-point, and the second derivative is also zero or very small. This condition (d²B/dx'² = 0) is characteristic of Helmholtz coils, ensuring a highly uniform field.

Let's analyze the field at the mid-point, x' = R/2.
Distance from first coil: x₁ = R/2
Distance from second coil: x₂ = R - R/2 = R/2

So, at the mid-point, x₁ = x₂ = R/2. The contributions from both coils are equal:
B₁ = B₂ = (μ₀NIR²) / (2((R/2)² + R²)^(3/2))
B₁ = (μ₀NIR²) / (2(R²/4 + R²)^(3/2))
B₁ = (μ₀NIR²) / (2(5R²/4)^(3/2))
B₁ = (μ₀NIR²) / (2 * (5^(3/2) * R³ / 4^(3/2)))
B₁ = (μ₀NIR²) / (2 * (5√5 * R³ / 8))
B₁ = (μ₀NIR²) / (5√5 * R³ / 4)
B₁ = (4μ₀NI) / (5√5 * R)

The total magnetic field at the mid-point (B_mid) is B₁ + B₂ = 2B₁:
B_mid = 2 * (4μ₀NI) / (5√5 * R)
B_mid = (8μ₀NI) / (5√5 * R)

Now, let's calculate the numerical value:
√5 ≈ 2.236
5√5 ≈ 11.180
B_mid = (8 / 11.180) * (μ₀NI / R)
B_mid ≈ 0.7155 * (μ₀NI / R)
B_mid ≈ 0.72 * (μ₀NI / R)

**Proof of Uniformity (d²B/dx'² = 0 at mid-point):**
Let f(x') = 1/(x'² + R²)^(3/2). Then B_total = (μ₀NIR²/2) * [f(x') + f(R - x')].
We need to show that d²B_total/dx'² = 0 at x' = R/2.

Let's find the first derivative of f(x'):
f'(x') = d/dx' [(x'² + R²)^(-3/2)] = (-3/2)(x'² + R²)^(-5/2) * (2x') = -3x'(x'² + R²)^(-5/2)

Now, the second derivative:
f''(x') = d/dx' [-3x'(x'² + R²)^(-5/2)]
Using product rule: f''(x') = -3(x'² + R²)^(-5/2) - 3x' * (-5/2)(x'² + R²)^(-7/2) * (2x')
f''(x') = -3(x'² + R²)^(-5/2) + 15x'²(x'² + R²)^(-7/2)
f''(x') = -3(x'² + R²)^(-7/2) [ (x'² + R²) - 5x'²/5 ]  -> this step is wrong, should be -3(x'² + R²)^(-7/2) [ (x'² + R²) - 5x'² ]
f''(x') = -3(x'² + R²)^(-7/2) [ (x'² + R²) - 5x'² ] = -3(x'² + R²)^(-7/2) [ R² - 4x'² ]

Now, for B_total:
dB_total/dx' = (μ₀NIR²/2) * [f'(x') - f'(R - x')]
At x' = R/2, f'(R/2) = -3(R/2)( (R/2)² + R² )^(-5/2) = -3R/2 (5R²/4)^(-5/2)
f'(R - R/2) = f'(R/2). So, dB_total/dx' = 0 at x' = R/2.

d²B_total/dx'² = (μ₀NIR²/2) * [f''(x') + f''(R - x')]
At x' = R/2, we need to evaluate f''(R/2).
f''(R/2) = -3((R/2)² + R²)^(-7/2) [ R² - 4(R/2)² ]
f''(R/2) = -3(5R²/4)^(-7/2) [ R² - R² ]
f''(R/2) = -3(5R²/4)^(-7/2) * 0 = 0

Since f''(R/2) = 0, and f''(R - R/2) = f''(R/2) = 0, then d²B_total/dx'² = 0 at x' = R/2.

This shows that the magnetic field is highly uniform around the mid-point between the coils in a Helmholtz arrangement, as both the first and second derivatives of the field are zero at the mid-point.

Question 13

Q13. A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an

Accepted Answer

To prevent the coil from turning, a counter torque equal in magnitude and opposite in direction to the magnetic torque acting on the coil must be applied. The magnetic torque (τ) on a current-carrying coil in a magnetic field is given by τ = NIAB sinθ, where N is the number of turns, I is the current, A is the area of the coil, B is the magnetic field strength, and θ is the angle between the magnetic field and the normal to the plane of the coil.

Given:
Number of turns, N = 30
Radius of the coil, r = 8.0 cm = 0.08 m
Current, I = 6.0 A
Magnetic field, B = 1.0 T
Angle between the magnetic field and the normal to the coil, θ = 60°

First, calculate the area of the coil:
A = πr² = π (0.08 m)² = π (0.0064 m²) = 0.020106 m²

Now, calculate the magnetic torque:
τ = N I A B sinθ
τ = 30 × 6.0 A × 0.020106 m² × 1.0 T × sin(60°)
τ = 30 × 6.0 × 0.020106 × 1.0 × (√3 / 2)
τ = 180 × 0.020106 × 0.866
τ = 3.61908 × 0.866
τ = 3.136 Nm

The magnitude of the counter torque that must be applied to prevent the coil from turning is 3.1 Nm (rounded to two significant figures).

Question 14

Q14. Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, are placed in the same vertical plane containing the north-south direction. Coil X has 20 turns and ca

Accepted Answer

We need to calculate the magnetic field produced by each coil at their common centre and then find the vector sum. The magnetic field at the centre of a circular coil is given by B = (μ₀NI) / (2R).

Given:
For coil X:
Radius, Rₓ = 16 cm = 0.16 m
Number of turns, Nₓ = 20
Current, Iₓ = 16 A (anticlockwise when viewed from west)

For coil Y:
Radius, Rᵧ = 10 cm = 0.10 m
Number of turns, Nᵧ = 25
Current, Iᵧ = 18 A (clockwise when viewed from west)

Permeability of free space, μ₀ = 4π × 10⁻⁷ T m A⁻¹

**1. Magnetic field due to coil X (Bₓ):**
Bₓ = (μ₀NₓIₓ) / (2Rₓ)
Bₓ = (4π × 10⁻⁷ T m A⁻¹ × 20 × 16 A) / (2 × 0.16 m)
Bₓ = (4π × 10⁻⁷ × 320) / 0.32
Bₓ = (4π × 10⁻⁷ × 1000)
Bₓ = 4π × 10⁻⁴ T
Bₓ ≈ 1.256 × 10⁻³ T

Direction of Bₓ: An observer looking from the west sees the current as anticlockwise. Using the right-hand thumb rule, if the fingers curl in the direction of the current (anticlockwise), the thumb points towards the observer (west). So, Bₓ is directed towards the west.

**2. Magnetic field due to coil Y (Bᵧ):**
Bᵧ = (μ₀NᵧIᵧ) / (2Rᵧ)
Bᵧ = (4π × 10⁻⁷ T m A⁻¹ × 25 × 18 A) / (2 × 0.10 m)
Bᵧ = (4π × 10⁻⁷ × 450) / 0.20
Bᵧ = (4π × 10⁻⁷ × 2250)
Bᵧ = 9000π × 10⁻⁷ T
Bᵧ = 9π × 10⁻⁴ T
Bᵧ ≈ 2.827 × 10⁻³ T

Direction of Bᵧ: An observer looking from the west sees the current as clockwise. Using the right-hand thumb rule, if the fingers curl in the direction of the current (clockwise), the thumb points away from the observer (east). So, Bᵧ is directed towards the east.

**3. Net magnetic field (B_net):**
Since Bₓ is towards the west and Bᵧ is towards the east, they are in opposite directions. We take the direction towards the east as positive and towards the west as negative.
B_net = Bᵧ - Bₓ (since Bᵧ > Bₓ)
B_net = 9π × 10⁻⁴ T - 4π × 10⁻⁴ T
B_net = 5π × 10⁻⁴ T
B_net ≈ 5 × 3.14159 × 10⁻⁴ T
B_net ≈ 1.5708 × 10⁻³ T

Since Bᵧ (east) is greater than Bₓ (west), the net magnetic field is directed towards the east.

Magnitude of net magnetic field = 1.57 × 10⁻³ T
Direction of net magnetic field = Towards the East.

Question 15

Q15. A magnetic field of 100 G (1 G = 10⁻⁴ T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10⁻³ m². The maximum current carry

Accepted Answer

To produce a uniform magnetic field in a specific region, a solenoid is an ideal choice. The magnetic field inside a long solenoid is given by B = μ₀nI, where B is the magnetic field strength, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.

Given:
Required magnetic field, B = 100 G = 100 × 10⁻⁴ T = 10⁻² T
Maximum current, I_max = 15 A
Maximum turns per unit length, n_max = 1000 turns/m
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m A⁻¹

We need to design a solenoid that can produce B = 10⁻² T using the available wire and current capacity.

From the formula B = μ₀nI, we can determine the required product of nI:
nI = B / μ₀
nI = (10⁻² T) / (4π × 10⁻⁷ T m A⁻¹)
nI = (1 / (4π)) × 10⁵ A/m
nI ≈ 0.079577 × 10⁵ A/m
nI ≈ 7957.7 A/m

Now, let's consider the constraints:

**Option 1: Using maximum current (I = 15 A)**
If we use the maximum current I = 15 A, then the required number of turns per unit length (n) would be:
n = (nI) / I = 7957.7 A/m / 15 A
n ≈ 530.5 turns/m
This value (530.5 turns/m) is well within the maximum winding capacity of 1000 turns/m. So, this is a feasible design.

**Option 2: Using maximum turns per unit length (n = 1000 turns/m)**
If we use the maximum turns per unit length n = 1000 turns/m, then the required current (I) would be:
I = (nI) / n = 7957.7 A/m / 1000 turns/m
I ≈ 7.9577 A
This current (7.96 A) is well within the maximum current carrying capacity of 15 A. This is also a feasible design.

**Suggested Design Particulars:**
Let's choose a design that uses a current less than the maximum capacity for safety and efficiency, and a reasonable number of turns per unit length.

We can choose a current, say, I = 10 A (which is less than 15 A).
Then, the required number of turns per unit length would be:
n = (nI) / I = 7957.7 A/m / 10 A
n ≈ 795.77 turns/m
This is less than the maximum winding capacity of 1000 turns/m.

To achieve a uniform field over a linear dimension of 10 cm, the length of the solenoid should be significantly greater than its diameter. Let's assume a diameter of about 5 cm (radius 2.5 cm) to fit the cross-sectional area requirement (Area ≈ πr² ≈ 3.14 × (0.025)² ≈ 0.00196 m² ≈ 10⁻³ m²).

Let's choose a length for the solenoid, say, L = 50 cm = 0.5 m (which is much greater than the 10 cm region where uniformity is required and also much greater than the assumed diameter).

Total number of turns, N = n × L = 795.77 turns/m × 0.5 m ≈ 398 turns.

**Summary of Suggested Design Particulars:**
*   **Type of Coil:** Solenoid (air-filled core)
*   **Magnetic Field Strength (B):** 100 G (10⁻² T)
*   **Current (I):** 10 A (within 15 A limit)
*   **Turns per unit length (n):** Approximately 796 turns/m (within 1000 turns/m limit)
*   **Length of Solenoid (L):** Approximately 50 cm (0.5 m) (to ensure uniformity over 10 cm region)
*   **Total Number of Turns (N):** Approximately 398 turns
*   **Diameter of Solenoid:** Approximately 5 cm (radius 2.5 cm) (to provide required cross-sectional area of ~10⁻³ m²)

This design ensures the required magnetic field strength and uniformity within the given constraints.

Question 16

Q16. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the (a) t

Accepted Answer

Given:
Number of turns, N = 20
Radius of coil, r = 10 cm = 0.1 m
Magnetic field, B = 0.10 T
Current in the coil, I = 5.0 A
Cross-sectional area of wire, A_wire = 10⁻⁵ m²
Free electron density, n = 10²⁹ m⁻³

(a) Total torque on the coil:
Since the magnetic field is normal to the plane of the coil, the angle between the magnetic moment vector (μ) and the magnetic field vector (B) is θ = 0° (or 180°). The torque τ is given by τ = μ × B = NIAB sinθ. Since sin 0° = 0, the total torque on the coil is zero.
τ = 0 Nm

(b) Total force on the coil:
For a current loop placed in a uniform magnetic field, the net magnetic force on the entire loop is always zero. This is because for every segment of the wire experiencing a force in one direction, there is a corresponding segment experiencing an equal and opposite force. Therefore, the total force on the coil is zero.
F_total = 0 N

(c) Average force on each electron in the coil due to the magnetic field:
First, we need to find the drift velocity (v_d) of the electrons. The current (I) is related to the drift velocity by the formula I = n A_wire e v_d, where e is the charge of an electron (1.6 × 10⁻¹⁹ C).
v_d = I / (n A_wire e)
v_d = 5.0 A / (10²⁹ m⁻³ × 10⁻⁵ m² × 1.6 × 10⁻¹⁹ C)
v_d = 5.0 / (1.6 × 10⁵) = 3.125 × 10⁻⁵ m/s

The magnetic force on a single electron is given by the Lorentz force formula, F_e = e (v_d × B). Since the magnetic field is normal to the plane of the coil, the drift velocity of electrons (which constitutes the current) is tangential to the coil. Thus, the drift velocity vector is perpendicular to the magnetic field vector (B) at every point along the wire (B is normal to the plane, v_d is in the plane). Therefore, the angle between v_d and B is 90°.
F_e = e v_d B sin 90°
F_e = e v_d B
F_e = (1.6 × 10⁻¹⁹ C) × (3.125 × 10⁻⁵ m/s) × (0.10 T)
F_e = 5.0 × 10⁻²⁵ N

Final Answer:
(a) Total torque on the coil = 0 Nm
(b) Total force on the coil = 0 N
(c) Average force on each electron = 5.0 × 10⁻²⁵ N

Question 17

Q17. A magnetic field of 100 G (1 G = 10⁻⁴ T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10⁻³ m². The maximum current carry

Accepted Answer

Given:
Required magnetic field, B = 100 G = 100 × 10⁻⁴ T = 10⁻² T
Maximum current, I_max = 15 A
Maximum number of turns per unit length, n_max = 1000 turns/m
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m/A

For a long solenoid, the magnetic field inside is given by B = μ₀ n I.
We need to design a solenoid such that B = 10⁻² T.

We can choose values for n and I such that their product (nI) gives the required magnetic field.
From B = μ₀ n I, we have nI = B / μ₀
nI = (10⁻² T) / (4π × 10⁻⁷ T m/A)
nI = (10⁵) / (4π) ≈ 7957.75 A/m

Now, we need to choose n and I such that n ≤ n_max (1000 turns/m) and I ≤ I_max (15 A), and their product is approximately 7957.75 A/m.

Let's try to use the maximum allowed number of turns per unit length, n = 1000 turns/m.
Then, I = (nI) / n = 7957.75 A/m / 1000 turns/m = 7.95775 A.
This current (7.95775 A) is less than the maximum current carrying capacity (15 A), so this is a feasible option.

Design Particulars:
1.  **Number of turns per unit length (n):** We can choose n = 1000 turns/m (maximum allowed). 2.  **Current (I):** With n = 1000 turns/m, the required current would be I = 7.96 A (approximately). This is well within the maximum current capacity of 15 A. 3.  **Length of the solenoid:** To ensure a uniform field over a linear dimension of 10 cm, the length of the solenoid should be significantly larger than 10 cm. A length of at least 50 cm (0.5 m) to 1 m would be appropriate to achieve good uniformity in the central region. 4.  **Radius of the solenoid:** The area of cross-section is about 10⁻³ m². If the cross-section is circular, then πr² ≈ 10⁻³ m², which gives r ≈ √(10⁻³ / π) ≈ 0.0178 m ≈ 1.8 cm. This radius is reasonable for a solenoid. The radius does not directly affect the magnetic field strength in a long solenoid, but it defines the volume over which the field is uniform. 5.  **Core material:** Non-ferromagnetic (as specified). Air core or vacuum core.  Suggested Design Particulars:
*   **Number of turns per unit length (n):** 1000 turns/m
*   **Current (I):** 8 A (rounded up slightly for practical purposes, still within 15 A limit)
*   **Length of the solenoid (L):** At least 0.5 m (e.g., 50 cm) to ensure uniformity over 10 cm.
*   **Radius of the solenoid (r):** Approximately 1.8 cm (corresponding to an area of 10⁻³ m²).

Let's verify the magnetic field with these parameters:
B = μ₀ n I = (4π × 10⁻⁷ T m/A) × (1000 turns/m) × (8 A)
B = 32π × 10⁻⁴ T ≈ 1.005 × 10⁻² T ≈ 100.5 G.
This is very close to the required 100 G.

Final Answer:
Appropriate design particulars for the solenoid:
*   Number of turns per unit length (n): 1000 turns/m
*   Current (I): 8 A
*   Length of the solenoid: 0.5 m (or more)
*   Radius of the solenoid: ~1.8 cm (to provide the specified cross-sectional area)

Question 18

Q18. Find the magnetic field at the centre of a circular coil of 100 turns, radius 10 cm, carrying a current of 0.4 A.

Accepted Answer

The magnetic field at the centre of a circular coil is given by the formula B = (μ₀ * N * I) / (2 * R). Substituting the given values, the magnetic field at the centre of the coil is 2.512 × 10⁻⁴ T.

Question 19

Q19. A long straight wire carries a current of 35 A. What is the magnitude of the magnetic field B at a point 20 cm from the wire?

Accepted Answer

The magnetic field at a point due to a long straight current-carrying wire is given by B = (μ₀ * I) / (2 * π * r). Substituting the given values, the magnitude of the magnetic field at 20 cm from the wire is 3.5 × 10⁻⁵ T.

Question 20

Q20. A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the (a) t

Accepted Answer

This problem involves calculating torque on a current loop, total force on a current loop in a uniform magnetic field, and the average force on individual electrons using drift velocity.

(a) The total torque on the coil is 0 N m.
(b) The total force on the coil is 0 N.
(c) The average force on each electron in the coil due to the magnetic field is approximately 5.0 × 10⁻²⁵ N.

Question 21

Q21. A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. (a) W

Accepted Answer

The problem involves the equilibrium of forces acting on a current-carrying conductor in a magnetic field. The forces are gravitational force (weight) acting downwards, and magnetic force (due to current in magnetic field) which can act upwards or downwards depending on the direction of current and magnetic field.

**(a) Magnetic field for zero tension:**
For the tension in the wires to be zero, the upward magnetic force on the rod must exactly balance its downward weight.
Given:
Length of the rod, L = 0.45 m
Mass of the rod, m = 60 g = 0.060 kg
Current in the rod, I = 5.0 A
Acceleration due to gravity, g = 9.8 m s⁻²

Weight of the rod, W = mg
Magnetic force on the rod, F_m = I L B sinθ
Since the magnetic field is normal to the conductor, θ = 90°, so sinθ = 1.
F_m = I L B

For zero tension, F_m = W
I L B = mg
B = mg / (I L)
B = (0.060 kg × 9.8 m s⁻²) / (5.0 A × 0.45 m)
B = 0.588 / 2.25
B = 0.2613 T

Rounding to two significant figures (as per given current and length), B ≈ 0.26 T.

The direction of the magnetic field should be such that the magnetic force (F = I(L x B)) is directed upwards. If the current is, say, from left to right, and the magnetic force is to be upwards, then by Fleming's Left Hand Rule, the magnetic field must be directed horizontally outwards (perpendicular to the rod and upwards force).

**(b) Total tension if current direction is reversed:**
If the direction of the current is reversed, while keeping the magnetic field direction the same, the direction of the magnetic force will also reverse. Now, both the magnetic force and the weight of the rod will act in the downward direction.

Total downward force = Weight + Magnetic force
Total tension, T = mg + I L B
From part (a), we know that I L B = mg (since this was the condition for zero tension).
So, T = mg + mg
T = 2 mg
T = 2 × (0.060 kg × 9.8 m s⁻²)
T = 2 × 0.588 N
T = 1.176 N

Rounding to two significant figures, T ≈ 1.2 N.

Question 22

Q22. The wires which connect the battery to the starting motor of an automobile carry a current of 300 A (for a short time). What is the force per unit length between the wires if they

Accepted Answer

This problem deals with the force between two parallel current-carrying conductors. The direction of the force (attractive or repulsive) depends on the relative directions of the currents.

Given:
Current, I = 300 A (This is the current in each wire, assuming the return path carries the same current)
Length of the wires, L = 70 cm = 0.70 m
Distance between the wires, d = 1.5 cm = 0.015 m
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m A⁻¹

**1. Calculate Force per Unit Length:**
The formula for the force per unit length (F/L) between two parallel current-carrying wires is:
F/L = (μ₀ I₁ I₂) / (2π d)
Since the wires connect a battery to a motor, one wire carries current from the battery to the motor, and the other wire carries current from the motor back to the battery. Therefore, the currents in the two wires are equal in magnitude (I₁ = I₂ = I = 300 A) but flow in opposite directions.

F/L = (4π × 10⁻⁷ T m A⁻¹ × 300 A × 300 A) / (2π × 0.015 m)
F/L = (2 × 10⁻⁷ × 300 × 300) / 0.015 N m⁻¹
F/L = (2 × 10⁻⁷ × 90000) / 0.015 N m⁻¹
F/L = (180000 × 10⁻⁷) / 0.015 N m⁻¹
F/L = (0.018) / 0.015 N m⁻¹
F/L = 1.2 N m⁻¹

**2. Determine the Nature of the Force:**
When currents in two parallel conductors flow in opposite directions, the magnetic fields they produce interact in such a way that the wires repel each other.

Therefore, the force is repulsive.

Question 23

Q23. A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the n

Accepted Answer

This problem involves calculating the magnetic force on a current-carrying wire in a uniform magnetic field, considering different orientations and shapes of the wire. The key formula is F = I (L x B) or F = I L B sinθ, where L is the effective length of the wire inside the magnetic field.

Given:
Magnetic field, B = 1.5 T (East to West)
Radius of cylindrical region, R = 10.0 cm = 0.10 m
Current, I = 7.0 A

**Part (a): The wire intersects the axis.**
When the wire passes through the cylindrical region intersecting its axis, the length of the wire inside the magnetic field is equal to the diameter of the cylindrical region.
Effective length, L_eff = 2R = 2 × 0.10 m = 0.20 m
The current flows North to South, and the magnetic field is East to West. These directions are perpendicular to each other (θ = 90°).

Magnitude of force, F = I L_eff B sinθ
F = 7.0 A × 0.20 m × 1.5 T × sin(90°)
F = 7.0 × 0.20 × 1.5 × 1 N
F = 2.1 N

Direction of force: Using Fleming's Left Hand Rule (or F = I(L x B)): Current (I) is North to South, Magnetic field (B) is East to West. Pointing the forefinger (B) towards West and middle finger (I) towards South, the thumb (F) points vertically downwards.
Therefore, the force is 2.1 N, vertically downwards.

**Part (b): The wire is turned from N-S to north-east-south-west direction.**
In this case, the wire still passes through the cylindrical region intersecting its axis, so the effective length inside the field remains the same.
Effective length, L_eff = 2R = 0.20 m
The current direction is North-East to South-West. The magnetic field is East to West. The angle (θ) between the current direction (NE-SW) and the magnetic field direction (E-W) is 45°.

Magnitude of force, F = I L_eff B sinθ
F = 7.0 A × 0.20 m × 1.5 T × sin(45°)
F = 7.0 × 0.20 × 1.5 × (1/√2) N
F = 2.1 × (1/√2) N
F = 2.1 / 1.414 N
F ≈ 1.485 N
Rounding to two significant figures, F ≈ 1.5 N.

Direction of force: Current (I) is NE-SW, Magnetic field (B) is E-W. Using Fleming's Left Hand Rule: Pointing the forefinger (B) towards West and middle finger (I) towards SW, the thumb (F) points vertically downwards. The direction of force remains vertically downwards, as both current and magnetic field are in the horizontal plane.
Therefore, the force is approximately 1.5 N, vertically downwards.

**Part (c): The wire in (a) is replaced by a semicircular wire of the same length that also intersects the axis.**
For a current-carrying wire of arbitrary shape in a uniform magnetic field, the magnetic force depends only on the current, the magnetic field, and the displacement vector (straight line distance) between the entry and exit points of the wire in the field. This is often referred to as the effective length.

In this case, the semicircular wire starts at one edge of the cylindrical region and ends at the opposite edge, intersecting the axis. The displacement vector between its entry and exit points is the diameter of the cylindrical region.
Effective length, L_eff = 2R = 0.20 m
The direction of this effective length vector is North to South (same as the straight wire in part (a)). The magnetic field is East to West. So, the angle between the effective length and the magnetic field is 90°.

Magnitude of force, F = I L_eff B sinθ
F = 7.0 A × 0.20 m × 1.5 T × sin(90°)
F = 7.0 × 0.20 × 1.5 × 1 N
F = 2.1 N

Direction of force: The direction of the force is determined by the effective length vector (N-S) and the magnetic field (E-W). Similar to part (a), the force will be vertically downwards.
Therefore, the force is 2.1 N, vertically downwards.

Question 24

Q24. A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires. (a) W

Accepted Answer

The problem involves the equilibrium of forces on a current-carrying conductor in a magnetic field. The forces acting on the rod are its weight (downwards), the tension in the wires (upwards), and the magnetic force (either upwards or downwards depending on the direction of current and magnetic field).

**(a) Magnetic field for zero tension:**
For the tension in the wires to be zero, the upward magnetic force must exactly balance the downward gravitational force (weight) of the rod.
Given:
Length of the rod, L = 0.45 m
Mass of the rod, m = 60 g = 0.060 kg
Current in the rod, I = 5.0 A
Acceleration due to gravity, g = 9.8 m/s²

Weight of the rod, W = mg = 0.060 kg × 9.8 m/s² = 0.588 N

The magnetic force on a current-carrying conductor in a magnetic field is given by F_m = I(L × B). Since the magnetic field is normal to the conductor, the magnitude of the force is F_m = BIL.

For zero tension, F_m = W
BIL = mg
B = mg / (IL)
B = 0.588 N / (5.0 A × 0.45 m)
B = 0.588 / 2.25 T
B ≈ 0.261 T

To have an upward magnetic force, by Fleming's Left-Hand Rule, if the current is, say, from left to right, and the magnetic field is vertically upwards, the force would be into the page. If the current is from left to right, and the magnetic field is into the page, the force would be upwards. So, the magnetic field must be perpendicular to the rod and in a direction such that the force BIL acts upwards.

**(b) Total tension if current direction is reversed:**
If the direction of the current is reversed, while keeping the magnetic field (B = 0.261 T) in the same direction, the direction of the magnetic force F_m will also reverse. Now, the magnetic force will act downwards, in the same direction as the weight of the rod.

Total downward force = Weight + Magnetic force
F_total = mg + BIL

We know mg = 0.588 N
And BIL = 0.588 N (from part a, as it balanced the weight)

So, F_total = 0.588 N + 0.588 N = 1.176 N

This total downward force must be balanced by the tension in the wires.
Therefore, the total tension in the wires = 1.176 N.

Summary:
(a) Magnetic field B ≈ 0.261 T, directed perpendicular to the rod such that the magnetic force acts upwards.
(b) Total tension in the wires = 1.176 N.

Question 25

Q25. The wires which connect the battery to the starting motor of an automobile carry a current of 300 A (for a short time). What is the force per unit length between the wires if they

Accepted Answer

This problem involves calculating the force between two parallel current-carrying conductors. The wires connecting the battery to the starting motor carry current in opposite directions (one wire carries current from the battery to the motor, and the other carries it back from the motor to the battery, completing the circuit).

Given:
Current, I = 300 A
Length of the wires, L = 70 cm = 0.70 m
Distance between the wires, r = 1.5 cm = 0.015 m

The force per unit length between two parallel current-carrying wires is given by the formula:
F/L = (μ₀ * I₁ * I₂) / (2πr)

In this case, both wires carry the same current, so I₁ = I₂ = I = 300 A.
μ₀ (permeability of free space) = 4π × 10⁻⁷ T m/A

Substitute the values into the formula:
F/L = (4π × 10⁻⁷ T m/A × 300 A × 300 A) / (2π × 0.015 m)
F/L = (2 × 10⁻⁷ × 300 × 300) / 0.015 N/m
F/L = (2 × 10⁻⁷ × 90000) / 0.015 N/m
F/L = (180000 × 10⁻⁷) / 0.015 N/m
F/L = 0.018 / 0.015 N/m
F/L = 1.2 N/m

**Nature of the force:**
Since the wires connect the battery to the motor, one wire carries current from the battery to the motor, and the other wire carries current back from the motor to the battery. Therefore, the currents in the two wires flow in opposite directions.
When currents in two parallel wires flow in opposite directions, the force between them is repulsive.

Thus, the force per unit length between the wires is 1.2 N/m, and the force is repulsive.

Question 26

Q26. A current-carrying circular loop of radius R is placed in the x-y plane with centre at the origin. For a small magnetic dipole moment, find the magnetic field at a point (0, 0, z)

Accepted Answer

To find the magnetic field at a point (0, 0, z) on the axis of the loop, considering the loop as a small magnetic dipole, we use the magnetic dipole approximation. The magnetic dipole moment of the circular loop is given by `m = I A = I πR²`, where `I` is the current and `R` is the radius of the loop. Since the loop is in the x-y plane, its magnetic dipole moment vector `m` is directed along the z-axis (assuming the current is anti-clockwise when viewed from positive z-axis, by the right-hand rule). So, `m = m k̂`.

The point (0, 0, z) lies on the z-axis, which is the axial line for this magnetic dipole. For a small magnetic dipole moment, we assume that the point of observation is far away from the loop, i.e., `z >> R`.

The magnetic field `B` at an axial point of a magnetic dipole is given by:
`B_axial = (μ₀ / 4π) * (2m / z³)`

Substituting `m = I πR²`:
`B_axial = (μ₀ / 4π) * (2 I πR² / z³)`

The direction of the magnetic field at an axial point is along the direction of the magnetic dipole moment. Therefore, the magnetic field `B` at (0, 0, z) is:
`B = (μ₀ / 4π) * (2 I πR² / z³) k̂`

Where:
*   `μ₀` is the permeability of free space.
*   `I` is the current in the loop.
*   `R` is the radius of the loop.
*   `z` is the distance of the point from the center of the loop along the axis.
*   `k̂` is the unit vector along the positive z-axis.

Question 27

Q27. A current-carrying circular loop of radius R is placed in the x-y plane with centre at the origin. For a small magnetic dipole moment, find the magnetic field at a point (x, 0, 0)

Accepted Answer

To find the magnetic field at a point (x, 0, 0), considering the loop as a small magnetic dipole, we use the magnetic dipole approximation. The magnetic dipole moment of the circular loop is given by `m = I A = I πR²`, where `I` is the current and `R` is the radius of the loop. Since the loop is in the x-y plane, its magnetic dipole moment vector `m` is directed along the z-axis (assuming the current is anti-clockwise when viewed from positive z-axis, by the right-hand rule). So, `m = m k̂`.

The point (x, 0, 0) lies on the x-axis. This point is in the x-y plane, which is the equatorial plane for the magnetic dipole `m` (since `m` is along the z-axis). For a small magnetic dipole moment, we assume that the point of observation is far away from the loop, i.e., `x >> R`.

The magnetic field `B` at an equatorial point of a magnetic dipole is given by:
`B_equatorial = (μ₀ / 4π) * (m / x³)`

Substituting `m = I πR²`:
`B_equatorial = (μ₀ / 4π) * (I πR² / x³)`

The direction of the magnetic field at an equatorial point is opposite to the direction of the magnetic dipole moment. Therefore, the magnetic field `B` at (x, 0, 0) is:
`B = - (μ₀ / 4π) * (I πR² / x³) k̂`

Where:
*   `μ₀` is the permeability of free space.
*   `I` is the current in the loop.
*   `R` is the radius of the loop.
*   `x` is the distance of the point from the center of the loop along the x-axis.
*   `k̂` is the unit vector along the positive z-axis (and `-k̂` is along the negative z-axis).

Question 28

Q28. A current-carrying circular loop of radius R is placed in the x-y plane with centre at the origin. For a small magnetic dipole moment, find the magnetic field at a point (0, y, 0)

Accepted Answer

To find the magnetic field at a point (0, y, 0), considering the loop as a small magnetic dipole, we use the magnetic dipole approximation. The magnetic dipole moment of the circular loop is given by `m = I A = I πR²`, where `I` is the current and `R` is the radius of the loop. Since the loop is in the x-y plane, its magnetic dipole moment vector `m` is directed along the z-axis (assuming the current is anti-clockwise when viewed from positive z-axis, by the right-hand rule). So, `m = m k̂`.

The point (0, y, 0) lies on the y-axis. This point is in the x-y plane, which is the equatorial plane for the magnetic dipole `m` (since `m` is along the z-axis). For a small magnetic dipole moment, we assume that the point of observation is far away from the loop, i.e., `y >> R`.

The magnetic field `B` at an equatorial point of a magnetic dipole is given by:
`B_equatorial = (μ₀ / 4π) * (m / y³)`

Substituting `m = I πR²`:
`B_equatorial = (μ₀ / 4π) * (I πR² / y³)`

The direction of the magnetic field at an equatorial point is opposite to the direction of the magnetic dipole moment. Therefore, the magnetic field `B` at (0, y, 0) is:
`B = - (μ₀ / 4π) * (I πR² / y³) k̂`

Where:
*   `μ₀` is the permeability of free space.
*   `I` is the current in the loop.
*   `R` is the radius of the loop.
*   `y` is the distance of the point from the center of the loop along the y-axis.
*   `k̂` is the unit vector along the positive z-axis (and `-k̂` is along the negative z-axis).

Chapter 4: Moving Charges and Magnetism