Q: Q13. At a certain location in Africa, a compass points 12° West of the geographic North. The North tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian po

Given:\n1. Magnetic declination (α) = 12° West of geographic North.\n2. Angle of dip (δ) = 60° (North tip points 60° above the horizontal in the magnetic meridian).\n3. Horizontal component of Earth's magnetic field (B_H) = 0.16 G.\n\n**Part 1: Magnitude of Earth's magnetic field (B)**\nWe know the relationship between the total Earth's magnetic field (B), its horizontal component (B_H), and the angle of dip (δ):\nB_H = B cos δ\n\nTherefore, B = B_H / cos δ\nB = 0.16 G / cos(60°)\nB = 0.16 G / 0.5\nB = 0.32 G\n\n**Part 2: Direction of Earth's magnetic field**\nThe direction of the Earth's magnetic field is specified by two angles:\n1. **Magnetic Declination (α):** This is the angle between the geographic meridian and the magnetic meridian. It is given as 12° West of the geographic North.\n2. **Magnetic Dip (δ):** This is the angle that the total magnetic field makes with the horizontal in the magnetic meridian. It is given as 60° above the horizontal.\n\nSo, the Earth's magnetic field at the location has a magnitude of **0.32 G** and its direction is **60° above the horizontal in the magnetic meridian, which itself is 12° West of the geographic North.**

Question 1

Q1. Answer the following questions:
(a) Why does a freely suspended magnet (or a magnetic needle) align itself in a particular direction?
(b) How does the magnetic field of the Earth

Accepted Answer

(a) A freely suspended magnet (or a magnetic needle) aligns itself in a particular direction (approximately North-South) because it experiences a torque due to the Earth's magnetic field. The Earth itself acts like a giant magnet, producing a magnetic field. When a magnet is suspended freely, its magnetic dipole moment tends to align itself with the direction of the external magnetic field (Earth's magnetic field) to achieve a state of minimum potential energy.

(b) The Earth's magnetic field protects us from cosmic rays, which are high-energy charged particles (protons, alpha particles, etc.) originating from outer space. When these charged particles approach the Earth, they encounter the Earth's magnetic field. According to the Lorentz force (F = q(v x B)), a force acts on these moving charged particles perpendicular to both their velocity (v) and the magnetic field (B). This force causes the charged particles to deflect from their original path, often trapping them in regions like the Van Allen belts or diverting them towards the magnetic poles. This deflection prevents most of these harmful particles from reaching the Earth's surface, thus protecting life.

(c) A magnetic field line is a continuous curve without breaks because magnetic monopoles do not exist. Magnetic field lines always form closed loops, originating from the North pole and ending at the South pole outside the magnet, and continuing from the South pole to the North pole inside the magnet. If magnetic monopoles existed, field lines could originate from a positive monopole and terminate at a negative monopole, having breaks. Since magnetic poles always exist in pairs (dipoles), field lines must be continuous.

(d) The angle of dip (or magnetic inclination) is the angle that the total intensity of the Earth's magnetic field makes with the horizontal plane at a given place. Its significance lies in providing information about the direction and strength of the Earth's magnetic field. It helps in determining the horizontal (BH) and vertical (BV) components of the Earth's magnetic field (BE) at a location: BV = BE sinδ and BH = BE cosδ, where δ is the angle of dip. The angle of dip is 0° at the magnetic equator and 90° at the magnetic poles, indicating the orientation of the Earth's magnetic field lines relative to the surface.

(e) The difference between the magnetic equator and the geographic equator is as follows:
*   **Geographic Equator:** It is an imaginary great circle on the Earth's surface, equidistant from the North and South geographic poles. It divides the Earth into the Northern and Southern Hemispheres. It is defined by the Earth's axis of rotation.
*   **Magnetic Equator:** It is an imaginary great circle on the Earth's surface where the Earth's magnetic field lines are parallel to the Earth's surface, and the angle of dip is 0°. It is not a fixed circle and does not coincide with the geographic equator. It is defined by the Earth's magnetic axis, which is tilted at an angle of approximately 11.3° with respect to the geographic axis.

Question 2

Q2. Answer the following questions:
(a) The Earth's magnetic field varies from place to place. Is it true or false?
(b) The magnetic field lines of the Earth are parallel to the surf

Accepted Answer

(a) **True.** The Earth's magnetic field is not uniform and varies significantly in magnitude and direction from one geographical location to another. This variation is due to the complex internal structure and dynamics of the Earth's core.

(b) **False.** At the magnetic poles, the Earth's magnetic field lines are perpendicular to the surface of the Earth (angle of dip is 90°).

(c) **False.** At the magnetic equator, the Earth's magnetic field lines are parallel to the surface of the Earth (angle of dip is 0°).

(d) **True.** At the magnetic equator, the Earth's magnetic field lines are parallel to the surface of the Earth, meaning the magnetic field is purely horizontal at these locations.

(e) **True.** At the magnetic poles, the Earth's magnetic field lines are perpendicular to the surface of the Earth, meaning the magnetic field is purely vertical at these locations.

Question 3

Q3. Answer the following questions:
(a) The Earth's magnetic field does not originate from a giant bar magnet located at the centre of the Earth. Why?
(b) The Earth's magnetic field

Accepted Answer

(a) The Earth's magnetic field does not originate from a giant bar magnet located at the center of the Earth for several reasons:
1.  **High Temperature:** The temperature inside the Earth's core is extremely high (estimated to be around 6000°C), far exceeding the Curie temperature of iron and nickel. Above the Curie temperature, ferromagnetic materials lose their magnetic properties and cannot sustain permanent magnetism.
2.  **Molten Core:** The Earth's outer core is in a molten state. A solid bar magnet cannot exist in this molten, high-temperature environment.
Instead, the Earth's magnetic field is believed to be generated by the 'dynamo effect', which involves the convection currents of molten metallic fluids (primarily iron and nickel) in the Earth's outer core, coupled with the Earth's rotation.

(b) **True.** The Earth's magnetic field is not static; it undergoes continuous changes over various timescales. These changes include variations in its strength, direction, and even the position of the magnetic poles. This phenomenon is known as secular variation.

(c) **True.** Paleomagnetic studies of rocks show clear evidence that the Earth's magnetic field has reversed its polarity numerous times throughout geological history. The North magnetic pole has become the South magnetic pole and vice-versa, with these reversals occurring irregularly over periods of thousands to millions of years.

(d) **False.** While the Earth's rotation is a crucial component of the dynamo effect, it is not the sole cause of the magnetic field. The rotation provides the Coriolis force, which organizes the convection currents of the molten material in the outer core into specific patterns. However, the primary source of the magnetic field is the movement of electrically conducting molten iron and nickel, not just the rotation itself.

(e) **True.** The most widely accepted theory for the origin of the Earth's magnetic field is the 'dynamo theory'. This theory attributes the magnetic field to the convection currents of electrically conducting molten iron and nickel in the Earth's outer core. These moving charges generate electric currents, which in turn produce the magnetic field.

Question 4

Q4. Answer the following questions:
(a) What is the difference between a diamagnetic and a paramagnetic substance?
(b) What is the difference between a paramagnetic and a ferromagnet

Accepted Answer

Question 5

Q5. A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude 4.5 × 10⁻² N m. What is the magnitude of the magne

Accepted Answer

The magnitude of the magnetic moment of the magnet is 0.36 N m/T (or A m²).

Question 6

Q6. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the equatorial line of the magnet (i.e., on

Accepted Answer

At a null point, the magnetic field due to the magnet is equal in magnitude and opposite in direction to the Earth's horizontal magnetic field. Since the angle of dip is zero, the Earth's total magnetic field is equal to its horizontal component (B_H). The null points are on the equatorial line of the magnet. We can use the formula for the magnetic field on the equatorial line of a short bar magnet to find its magnetic moment.

Question 7

Q7. A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the

Accepted Answer

We need to calculate the magnetic field strength and direction at two specific points: on the axial line and on the equatorial line of a short bar magnet, given its magnetic moment and the distance from its center.

Question 8

Q8. A solenoid with N turns, length l and radius r carries a current I. It is placed with its axis along the Z-axis. A coil of n turns and radius r' is placed such that it is coaxial w

Accepted Answer

To find the induced EMF in the coil when the current in the solenoid is reversed, we need to use Faraday's Law of Electromagnetic Induction. This involves calculating the initial and final magnetic flux through the coil and then determining the rate of change of flux.

Question 9

Q9. An iron rod of susceptibility 599 is subjected to a magnetising field of 1200 A m⁻¹. Find the permeability, magnetic susceptibility and magnetisation of the rod.

Accepted Answer

We are given the magnetic susceptibility and the magnetising field. We need to find the permeability and magnetisation of the iron rod. Note that magnetic susceptibility is already given, so we just need to state it.

Question 10

Q10. A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts as a bar magnet. What is its

Accepted Answer

A current-carrying solenoid behaves like a bar magnet because it produces a magnetic field pattern similar to that of a bar magnet. One end of the solenoid acts as a North pole and the other as a South pole. The magnetic moment associated with this solenoid can be calculated using its number of turns, current, and cross-sectional area.

Question 11

Q11. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°. If the dipole comes to stable equilibrium at an angle of 15° with th

Accepted Answer

Let the two magnetic fields be B₁ and B₂. The angle between them is θ = 60°. Let the magnetic moment of the dipole be M. At stable equilibrium, the magnetic dipole aligns itself with the resultant magnetic field (B_net). If the dipole makes an angle α = 15° with B₁, then the resultant field B_net also makes an angle of 15° with B₁. The angle B_net makes with B₂ will be (60° - 15°) = 45°.

Using the sine rule in the triangle formed by B₁, B₂, and B_net (or by resolving components):

From the sine rule for vectors:
B₁ / sin(angle between B_net and B₂) = B₂ / sin(angle between B_net and B₁)
B₁ / sin(45°) = B₂ / sin(15°)

B₁ / (1/√2) = B₂ / (sin(45°-30°))
B₁ / (1/√2) = B₂ / (sin45°cos30° - cos45°sin30°)
B₁ / (1/√2) = B₂ / ((1/√2)(√3/2) - (1/√2)(1/2))
B₁ / (1/√2) = B₂ / ((√3 - 1) / (2√2))

B₁ / B₂ = (1/√2) / ((√3 - 1) / (2√2))
B₁ / B₂ = 2 / (√3 - 1)

To rationalize the denominator:
B₁ / B₂ = 2(√3 + 1) / ((√3 - 1)(√3 + 1))
B₁ / B₂ = 2(√3 + 1) / (3 - 1)
B₁ / B₂ = 2(√3 + 1) / 2
B₁ / B₂ = √3 + 1

Therefore, the ratio of the magnitudes of the two fields is (√3 + 1) : 1.

Question 12

Q12. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its North tip pointing up at 22° with the horizontal. The horizontal component of the ear

Accepted Answer

The angle at which the North tip of the magnetic needle points up with the horizontal in the magnetic meridian is the angle of dip (δ).
Given:
Angle of dip, δ = 22°
Horizontal component of Earth's magnetic field, B_H = 0.35 G

We know the relationship between the Earth's total magnetic field (B), its horizontal component (B_H), and the angle of dip (δ):
B_H = B cos δ

To find the magnitude of the Earth's magnetic field (B):
B = B_H / cos δ

Substitute the given values:
B = 0.35 G / cos(22°)

Using cos(22°) ≈ 0.9272
B = 0.35 G / 0.9272
B ≈ 0.377 G

Therefore, the magnitude of the Earth's magnetic field at the place is approximately 0.377 G.

Question 13

Q13. At a certain location in Africa, a compass points 12° West of the geographic North. The North tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian po

Accepted Answer

Given:
1.  Magnetic declination (α) = 12° West of geographic North.
2.  Angle of dip (δ) = 60° (North tip points 60° above the horizontal in the magnetic meridian).
3.  Horizontal component of Earth's magnetic field (B_H) = 0.16 G.

**Part 1: Magnitude of Earth's magnetic field (B)**
We know the relationship between the total Earth's magnetic field (B), its horizontal component (B_H), and the angle of dip (δ):
B_H = B cos δ

Therefore, B = B_H / cos δ
B = 0.16 G / cos(60°)
B = 0.16 G / 0.5
B = 0.32 G

**Part 2: Direction of Earth's magnetic field**
The direction of the Earth's magnetic field is specified by two angles:
1.  **Magnetic Declination (α):** This is the angle between the geographic meridian and the magnetic meridian. It is given as 12° West of the geographic North.
2.  **Magnetic Dip (δ):** This is the angle that the total magnetic field makes with the horizontal in the magnetic meridian. It is given as 60° above the horizontal.

So, the Earth's magnetic field at the location has a magnitude of **0.32 G** and its direction is **60° above the horizontal in the magnetic meridian, which itself is 12° West of the geographic North.**

Question 14

Q14. A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the bar magnet at a distance of 10 cm from the centre of

Accepted Answer

Given:
Magnetic moment of the bar magnet, M = 0.48 J T⁻¹
Distance from the centre of the magnet, r = 10 cm = 0.10 m
Permeability of free space, μ₀ = 4π × 10⁻⁷ T m A⁻¹

**Part (a): Magnetic field on the axis**
For a short bar magnet, the magnetic field (B_axial) at a distance r from its centre on the axial line is given by:
B_axial = (μ₀ / 4π) * (2M / r³)

Substitute the values:
B_axial = (10⁻⁷ T m A⁻¹) * (2 * 0.48 J T⁻¹) / (0.10 m)³
B_axial = 10⁻⁷ * (0.96) / (0.001)
B_axial = 10⁻⁷ * 960
B_axial = 9.6 × 10⁻⁵ T

**Direction:** On the axial line, the magnetic field is in the same direction as the magnetic moment (from South pole to North pole of the magnet). If the magnet's North pole is pointing in a certain direction, the field on the axis beyond the North pole will be in that direction, and beyond the South pole, it will also be in that direction (i.e., along the axis, away from the North pole and towards the South pole, effectively in the direction of M).

**Part (b): Magnetic field on the equatorial lines (normal bisector)**
For a short bar magnet, the magnetic field (B_equatorial) at a distance r from its centre on the equatorial line is given by:
B_equatorial = (μ₀ / 4π) * (M / r³)

Substitute the values:
B_equatorial = (10⁻⁷ T m A⁻¹) * (0.48 J T⁻¹) / (0.10 m)³
B_equatorial = 10⁻⁷ * (0.48) / (0.001)
B_equatorial = 10⁻⁷ * 480
B_equatorial = 4.8 × 10⁻⁵ T

**Direction:** On the equatorial line, the magnetic field is in the direction opposite to the magnetic moment (from North pole to South pole of the magnet). If the magnetic moment M points from South to North, the field on the equatorial line will point from North to South.

**Summary:**
(a) On the axis: Magnitude = 9.6 × 10⁻⁵ T, Direction = Along the direction of the magnetic moment.
(b) On the equatorial line: Magnitude = 4.8 × 10⁻⁵ T, Direction = Opposite to the direction of the magnetic moment.

Question 15

Q15. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°. If the dipole comes to stable equilibrium at an angle of 15° with th

Accepted Answer

Let the two magnetic fields be B₁ and B₂. The angle between their directions is θ = 60°. The magnetic dipole comes to stable equilibrium at an angle α = 15° with the direction of B₁. In stable equilibrium, the net torque on the magnetic dipole is zero, meaning the dipole aligns itself with the resultant magnetic field. Therefore, the resultant magnetic field makes an angle of 15° with B₁.

Let M be the magnetic moment of the dipole. The torque exerted by a magnetic field B on a dipole M is given by τ = M × B, or τ = MB sinφ, where φ is the angle between M and B.

When the dipole is in stable equilibrium, it aligns with the resultant magnetic field. If the dipole makes an angle of 15° with B₁, then the angle it makes with B₂ must be (60° - 15°) = 45°. (Assuming B₂ is 'ahead' of B₁ by 60° and the resultant is between them).

For the dipole to be in equilibrium, the torques due to B₁ and B₂ must balance each other. This means the magnitude of the torque due to B₁ must be equal to the magnitude of the torque due to B₂.

Torque due to B₁: τ₁ = M B₁ sin(15°)
Torque due to B₂: τ₂ = M B₂ sin(45°)

For equilibrium, τ₁ = τ₂:
M B₁ sin(15°) = M B₂ sin(45°)
B₁ sin(15°) = B₂ sin(45°)

We need to find the ratio B₁/B₂:
B₁/B₂ = sin(45°) / sin(15°)

We know the values:
sin(45°) = 1/√2
sin(15°) = sin(45° - 30°) = sin(45°)cos(30°) - cos(45°)sin(30°)
= (1/√2)(√3/2) - (1/√2)(1/2)
= (√3 - 1) / (2√2)

Now, substitute these values into the ratio:
B₁/B₂ = (1/√2) / [(√3 - 1) / (2√2)]
B₁/B₂ = 2 / (√3 - 1)

To rationalize the denominator, multiply the numerator and denominator by (√3 + 1):
B₁/B₂ = 2(√3 + 1) / [(√3 - 1)(√3 + 1)]
B₁/B₂ = 2(√3 + 1) / (3 - 1)
B₁/B₂ = 2(√3 + 1) / 2
B₁/B₂ = √3 + 1

Using √3 ≈ 1.732:
B₁/B₂ = 1.732 + 1 = 2.732

The ratio of the magnitudes of the two fields is approximately 2.732.

Question 16

Q16. A place where the horizontal component of earth’s magnetic field is 0.3 G and the angle of dip is 30°. Find (a) the total magnetic field on the earth’s surface and (b) the vertical

Accepted Answer

Given:
Horizontal component of Earth's magnetic field, H = 0.3 G
Angle of dip, δ = 30°

We know that 1 Gauss (G) = 10⁻⁴ Tesla (T).
So, H = 0.3 × 10⁻⁴ T

(a) **Total magnetic field on the Earth's surface (B):**
The horizontal component (H) and the total magnetic field (B) are related by the angle of dip (δ) as:
H = B cos δ

Rearranging for B:
B = H / cos δ
B = (0.3 × 10⁻⁴ T) / cos(30°)
B = (0.3 × 10⁻⁴ T) / (√3 / 2)
B = (0.3 × 10⁻⁴ T) / (1.732 / 2)
B = (0.3 × 10⁻⁴ T) / 0.866
B ≈ 0.3464 × 10⁻⁴ T
B ≈ 0.3464 G

(b) **Vertical component of Earth's magnetic field (V):**
The vertical component (V) can be found using the total magnetic field (B) and the angle of dip (δ):
V = B sin δ

V = (0.3464 × 10⁻⁴ T) × sin(30°)
V = (0.3464 × 10⁻⁴ T) × (1/2)
V = 0.1732 × 10⁻⁴ T
V = 0.1732 G

Alternatively, V can be found using H and δ:
V = H tan δ
V = (0.3 × 10⁻⁴ T) × tan(30°)
V = (0.3 × 10⁻⁴ T) × (1/√3)
V = (0.3 × 10⁻⁴ T) / 1.732
V ≈ 0.1732 × 10⁻⁴ T
V ≈ 0.1732 G

Therefore:
(a) The total magnetic field on the Earth's surface is approximately 0.346 G.
(b) The vertical component of Earth's magnetic field is approximately 0.173 G.

Question 17

Q17. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the e

Accepted Answer

Given:
Angle of dip, δ = 22° (since the north tip points down at 22° with the horizontal in the magnetic meridian).
Horizontal component of Earth's magnetic field, H = 0.35 G.

We need to find the magnitude of the Earth's magnetic field, B.

First, convert the horizontal component from Gauss to Tesla:
H = 0.35 G = 0.35 × 10⁻⁴ T

The relationship between the horizontal component (H), the total magnetic field (B), and the angle of dip (δ) is:
H = B cos δ

To find B, we rearrange the formula:
B = H / cos δ

Substitute the given values:
B = (0.35 × 10⁻⁴ T) / cos(22°)

Calculate cos(22°):
cos(22°) ≈ 0.9272

Now, calculate B:
B = (0.35 × 10⁻⁴ T) / 0.9272
B ≈ 0.3775 × 10⁻⁴ T

Converting back to Gauss (optional, but good for comparison with given H):
B ≈ 0.3775 G

Therefore, the magnitude of the Earth's magnetic field at the place is approximately 0.378 G (or 0.378 × 10⁻⁴ T).

Question 18

Q18. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60° and one of the fields has a magnitude of 1.2 × 10⁻² T. If the dipole

Accepted Answer

The magnitude of the other magnetic field is approximately 4.39 × 10⁻³ T.

Question 19

Q19. A magnetic dipole of magnetic moment 1.45 J T⁻¹ is placed in a uniform magnetic field of magnitude 0.65 T, in such a way that its axis is aligned with the field.
(a) What is the a

Accepted Answer

(a) The amount of work done to turn the dipole axis 60° is 0.47125 J.
(b) The torque on the dipole in the new orientation is approximately 0.816 N m.

Question 20

Q20. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the cen

Accepted Answer

(a) The magnetic moment of the magnet is approximately 4.98 A m².
(b) If the magnet is turned around by 180°, the null points will be located on the equatorial line (normal bisector) at a distance of approximately 11.1 cm from the centre of the magnet.

Question 21

Q21. A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the

Accepted Answer

(a) On the axis: The magnitude of the magnetic field is 9.6 × 10⁻⁵ T. The direction is along the axis, in the direction of the magnetic moment (from the S-pole to the N-pole of the magnet).
(b) On the equatorial lines: The magnitude of the magnetic field is 4.8 × 10⁻⁵ T. The direction is along the equatorial line, opposite to the direction of the magnetic moment (from the N-pole to the S-pole of the magnet).

Question 22

Q22. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60° and one of the fields has a magnitude of 1.2 × 10⁻² T. If the dipole

Accepted Answer

Let the two magnetic fields be B₁ and B₂. Given B₁ = 1.2 × 10⁻² T. The angle between B₁ and B₂ is θ = 60°. The magnetic dipole comes to stable equilibrium at an angle α₁ = 15° with B₁. In stable equilibrium, the net torque on the dipole is zero, and the dipole moment aligns with the resultant magnetic field.

Let m be the magnetic moment of the dipole. The resultant magnetic field B_net makes an angle α₁ = 15° with B₁. The angle between B_net and B₂ will be α₂ = θ - α₁ = 60° - 15° = 45°.

For stable equilibrium, the dipole moment m aligns with the resultant magnetic field B_net. The components of B_net along and perpendicular to B₁ can be considered. Alternatively, we can use the sine rule for the vector triangle formed by B₁, B₂, and B_net.

Consider the vector addition of B₁ and B₂ to form B_net. Using the sine rule in the triangle formed by B₁, B₂, and B_net:
B₁ / sin(angle between B_net and B₂) = B₂ / sin(angle between B_net and B₁) = B_net / sin(angle between B₁ and B₂)

So, B₁ / sin(α₂) = B₂ / sin(α₁)
B₁ / sin(45°) = B₂ / sin(15°)

B₂ = B₁ * (sin(15°) / sin(45°))
B₂ = (1.2 × 10⁻² T) * (sin(15°) / sin(45°))

We know sin(15°) = sin(45° - 30°) = sin(45°)cos(30°) - cos(45°)sin(30°)
= (1/√2)(√3/2) - (1/√2)(1/2) = (√3 - 1) / (2√2) ≈ 0.2588
And sin(45°) = 1/√2 ≈ 0.7071

B₂ = (1.2 × 10⁻² T) * (0.2588 / 0.7071)
B₂ ≈ (1.2 × 10⁻² T) * 0.366
B₂ ≈ 0.4392 × 10⁻² T
B₂ ≈ 4.4 × 10⁻³ T

The magnitude of the other field is approximately 4.4 × 10⁻³ T.

Question 23

Q23. A magnetic dipole of magnetic moment 6.7 × 10⁻² A m² is placed in a uniform magnetic field of 0.06 T such that the dipole moment is aligned at 30° with the field. What is (a) the m

Accepted Answer

Given:
Magnetic moment of the dipole, m = 6.7 × 10⁻² A m²
Uniform magnetic field, B = 0.06 T
Angle between dipole moment and magnetic field, θ = 30°

**(a) Magnitude of the torque acting on the dipole:**
The formula for the torque (τ) acting on a magnetic dipole in a uniform magnetic field is given by:
τ = mB sinθ

Substituting the given values:
τ = (6.7 × 10⁻² A m²) × (0.06 T) × sin(30°)
τ = (6.7 × 10⁻² × 0.06) × (0.5) N m
τ = (0.402 × 10⁻²) × 0.5 N m
τ = 0.201 × 10⁻² N m
τ = 2.01 × 10⁻³ N m

**(b) Potential energy of the dipole in the field:**
The potential energy (U) of a magnetic dipole in a uniform magnetic field is given by:
U = -mB cosθ

Substituting the given values:
U = -(6.7 × 10⁻² A m²) × (0.06 T) × cos(30°)
U = -(6.7 × 10⁻² × 0.06) × (√3 / 2) J
U = -(0.402 × 10⁻²) × (0.866) J
U = -0.348 × 10⁻² J
U = -3.48 × 10⁻³ J

Therefore, (a) the magnitude of the torque acting on the dipole is 2.01 × 10⁻³ N m, and (b) the potential energy of the dipole in the field is -3.48 × 10⁻³ J.

Question 24

Q24. A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m² carrying a current of 4.0 A is suspended through its centre allowing it to turn in a horizontal plane

Accepted Answer

Given:
Number of turns, N = 2000
Area of cross-section, A = 1.6 × 10⁻⁴ m²
Current, I = 4.0 A
Uniform horizontal magnetic field, B = 7.5 × 10⁻² T
Angle between solenoid axis and magnetic field, θ = 30°
Moment of inertia, I_inertia = 2.5 × 10⁻⁵ kg m²

**(a) What is the magnetic moment associated with the solenoid?**
The magnetic moment (m) of a solenoid is given by the formula:
m = NIA
Where N is the number of turns, I is the current, and A is the area of cross-section.

Substituting the given values:
m = 2000 × 4.0 A × 1.6 × 10⁻⁴ m²
m = 8000 × 1.6 × 10⁻⁴ A m²
m = 12800 × 10⁻⁴ A m²
m = 1.28 A m²

**(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 30° with the axis of the solenoid?**

**Force:**
In a uniform magnetic field, the net force on a magnetic dipole (like a solenoid) is always zero. This is because the forces on opposite sides of the solenoid cancel each other out.
Therefore, the net force on the solenoid is **F = 0**.

**Torque:**
The torque (τ) acting on a magnetic dipole in a uniform magnetic field is given by:
τ = mB sinθ
Where m is the magnetic moment, B is the magnetic field, and θ is the angle between the magnetic moment vector (axis of solenoid) and the magnetic field.

Substituting the calculated magnetic moment and given values:
τ = (1.28 A m²) × (7.5 × 10⁻² T) × sin(30°)
τ = (1.28 × 7.5 × 10⁻²) × 0.5 N m
τ = (9.6 × 10⁻²) × 0.5 N m
τ = 4.8 × 10⁻² N m

**(c) What is the total torque on the solenoid due to the field if it is allowed to turn and align itself with the field?**
When the solenoid is allowed to turn and align itself with the magnetic field, it reaches a stable equilibrium position. In stable equilibrium, the magnetic moment vector (axis of the solenoid) becomes parallel to the magnetic field (θ = 0°).

At this position, the torque acting on the solenoid is:
τ = mB sin(0°)
τ = mB × 0
τ = 0

Therefore, the total torque on the solenoid when it aligns itself with the field is **zero**.

**(d) When does the solenoid experience a net force?**
A solenoid (or any magnetic dipole) experiences a net force only when it is placed in a **non-uniform magnetic field**. In a non-uniform field, the forces on different parts of the dipole do not cancel out, resulting in a net force. This force tends to move the dipole from a region of weaker field to a region of stronger field (if the dipole is aligned with the field gradient) or vice-versa.

**(e) If the solenoid is rotated by a small angle α from its stable equilibrium position, show that it executes simple harmonic motion. The moment of inertia of the solenoid about its axis of rotation is 2.5 × 10⁻⁵ kg m².**

**1. Torque acting on the solenoid:**
When the solenoid is rotated by a small angle α from its stable equilibrium position (where θ = 0°), the torque acting on it tends to restore it to the equilibrium position. The angle between the magnetic moment and the field is now α.
τ = -mB sinα
The negative sign indicates that the torque is a restoring torque, acting opposite to the angular displacement.

**2. Small angle approximation:**
For small angles, sinα ≈ α (where α is in radians).
So, τ ≈ -mBα

**3. Relation between torque and angular acceleration:**
From Newton's second law for rotational motion, torque is given by:
τ = I_inertia (d²α/dt²)
Where I_inertia is the moment of inertia and (d²α/dt²) is the angular acceleration.

**4. Equation of motion:**
Equating the two expressions for torque:
I_inertia (d²α/dt²) = -mBα
d²α/dt² = -(mB / I_inertia) α

**5. Comparison with SHM equation:**
This equation is in the form of the differential equation for simple harmonic motion:
d²α/dt² = -ω²α
Where ω is the angular frequency of oscillation.

Comparing the two equations, we get:
ω² = mB / I_inertia
ω = √(mB / I_inertia)

Since ω is a constant (m, B, and I_inertia are constants), the motion is simple harmonic.

**6. Calculation of angular frequency and time period (optional but good for completeness):**
m = 1.28 A m²
B = 7.5 × 10⁻² T
I_inertia = 2.5 × 10⁻⁵ kg m²

ω = √[(1.28 A m² × 7.5 × 10⁻² T) / (2.5 × 10⁻⁵ kg m²)]
ω = √[(9.6 × 10⁻²) / (2.5 × 10⁻⁵)] rad/s
ω = √[ (9.6 / 2.5) × 10³ ] rad/s
ω = √[3.84 × 10³] rad/s
ω = √[3840] rad/s
ω ≈ 61.97 rad/s

The time period of oscillation, T = 2π/ω
T = 2π / 61.97 s
T ≈ 0.101 s

Thus, the solenoid executes simple harmonic motion when rotated by a small angle from its stable equilibrium position.

Question 25

Q25. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°. If the dipole comes to stable equilibrium at an angle of 15° with on

Accepted Answer

The magnitude of the other magnetic field is approximately 4.39 × 10⁻³ T.

Chapter 5: Magnetism and Matter