Question 1

Q1. Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).

Accepted Answer

The direction of induced current is determined by Lenz's Law, which states that the induced current will flow in a direction that opposes the change in magnetic flux that caused it. We use the right-hand thumb rule to relate current direction to magnetic field direction.

**(a) Magnet moving towards a coil:**
As the North pole of the magnet moves towards the coil, the magnetic flux linked with the coil in the downward direction (towards the coil) increases. To oppose this increase, the induced current in the coil will create a North pole on the side facing the magnet. By the right-hand thumb rule, the current will flow **clockwise** when viewed from the magnet side (or from the right).

**(b) Magnet moving away from a coil:**
As the South pole of the magnet moves away from the coil, the magnetic flux linked with the coil in the upward direction (away from the coil) decreases. To oppose this decrease, the induced current in the coil will create a North pole on the side facing the magnet. By the right-hand thumb rule, the current will flow **clockwise** when viewed from the magnet side (or from the left).

**(c) Tapping key just closed:**
When the tapping key is just closed, the current in the primary coil (left) increases from zero. This creates a magnetic field in the primary coil, directed to the right. As this field increases, the magnetic flux linked with the secondary coil (right) to the right increases. To oppose this increase, the induced current in the secondary coil will create a magnetic field to the left. By the right-hand thumb rule, the current will flow from **'p' to 'q'**.

**(d) Rheostat setting being changed (decreasing current):**
When the rheostat setting is changed to decrease the current in the primary coil (left), the magnetic field produced by the primary coil (directed to the right) decreases. Consequently, the magnetic flux linked with the secondary coil (right) to the right decreases. To oppose this decrease, the induced current in the secondary coil will create a magnetic field to the right. By the right-hand thumb rule, the current will flow from **'q' to 'p'**.

**(e) Current in the wire decreasing:**
When the current in the straight wire (AB) decreases, the magnetic field produced by it (into the page below the wire, out of the page above the wire) decreases. For the rectangular loop, the magnetic flux into the page decreases. To oppose this decrease, the induced current in the loop will create a magnetic field into the page. By the right-hand thumb rule, the current will flow **clockwise**.

**(f) Loop moving into a region of magnetic field:**
As the loop moves into the region of the magnetic field (which is directed into the page), the magnetic flux linked with the loop (into the page) increases. To oppose this increase, the induced current in the loop will create a magnetic field out of the page. By the right-hand thumb rule, the current will flow **anticlockwise**.

Question 2

Q2. Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19: (a) A wire of irregular shape turning into a circular shape; (b) A circular l

Accepted Answer

Lenz's Law states that the direction of the induced current is such that it opposes the change in magnetic flux that produced it.

**(a) A wire of irregular shape turning into a circular shape:**
*   **Change in Flux:** When an irregular shaped wire (assuming it encloses a certain area) is pulled into a circular shape, its area increases. If there is an external magnetic field (e.g., perpendicular to the plane of the wire, say, into the page), the magnetic flux linked with the loop (into the page) will increase because the area is increasing.
*   **Opposing the Change:** To oppose this increase in inward magnetic flux, the induced current must create an outward magnetic field.
*   **Direction of Induced Current:** By the right-hand thumb rule, to produce an outward magnetic field, the induced current must flow in the **anticlockwise** direction.

**(b) A circular loop being deformed into a narrow straight wire:**
*   **Change in Flux:** When a circular loop is deformed into a narrow straight wire, its area decreases significantly (approaching zero). If there is an external magnetic field (e.g., perpendicular to the plane of the loop, say, into the page), the magnetic flux linked with the loop (into the page) will decrease because the area is decreasing.
*   **Opposing the Change:** To oppose this decrease in inward magnetic flux, the induced current must create an inward magnetic field.
*   **Direction of Induced Current:** By the right-hand thumb rule, to produce an inward magnetic field, the induced current must flow in the **clockwise** direction.

Question 3

Q3. A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.

Accepted Answer

To find the induced emf in the loop, we will first calculate the magnetic field inside the solenoid, then the change in magnetic flux through the loop, and finally apply Faraday's Law of electromagnetic induction.

Question 4

Q4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf

Accepted Answer

The induced emf in a conductor moving in a magnetic field (motional emf) is given by ε = Blv, where B is the magnetic field, l is the length of the conductor cutting the flux lines, and v is the velocity perpendicular to both B and l. The duration of the induced voltage depends on the width of the loop that crosses the field boundary.

Question 5

Q5. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact wi

Accepted Answer

When a metallic rod rotates in a uniform magnetic field, a motional emf is induced across its ends. Since the magnetic field is parallel to the axis of rotation, it is perpendicular to the velocity of every point on the rod. We can find the induced emf by integrating the motional emf across infinitesimal segments of the rod.

Question 6

Q6. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact wi

Accepted Answer

The emf developed between the centre and the ring is 100 V.

Question 7

Q7. A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s⁻¹ in a uniform horizontal magnetic field of magnitude 3.0 × 1

Accepted Answer

The maximum induced emf is 0.603 V. The average induced emf over a full cycle is 0 V. The maximum current is 0.0603 A. The average current over a full cycle is 0 A. The total power loss due to Joule heating is 0.0182 W. This power comes from the external mechanical agent rotating the coil.

Question 8

Q8. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s⁻¹ at right angles to the horizontal component of the earth’s magnetic field, 0.3

Accepted Answer

(a) The instantaneous value of the emf induced in the wire is 1.5 × 10⁻³ V.
(b) The direction of the induced emf is from West to East.
(c) The West end of the wire is at the higher electrical potential.

Question 9

Q9. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Accepted Answer

The change of flux linkage with the other coil is 30 Wb.

Question 10

Q10. A jet plane is travelling west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing, 25 m long, if the Earth’s magnetic field at the locat

Accepted Answer

The voltage difference developed between the ends of the wing is 3.125 V.

Question 11

Q11. A rectangular loop of sides 8 cm × 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf develop

Accepted Answer

The emf developed across the cut is due to motional emf, given by E = Bvl, where B is the magnetic field, v is the velocity, and l is the length of the side cutting the magnetic field lines.

Given:
Length of the longer side, L = 8 cm = 0.08 m
Length of the shorter side, w = 2 cm = 0.02 m
Magnetic field, B = 0.3 T
Velocity of the loop, v = 1 cm s⁻¹ = 0.01 m s⁻¹

(a) Velocity normal to the longer side:
When the velocity is normal to the longer side (L), the shorter side (w) is cutting the magnetic field lines. So, the effective length 'l' for emf induction is the shorter side, w.

(b) Velocity normal to the shorter side:
When the velocity is normal to the shorter side (w), the longer side (L) is cutting the magnetic field lines. So, the effective length 'l' for emf induction is the longer side, L.

Question 12

Q12. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact wi

Accepted Answer

When a metallic rod rotates in a uniform magnetic field, a motional emf is induced across its ends. For a rotating rod, the velocity of each segment is different, so we need to integrate to find the total emf.

Given:
Length of the rod, L = 1.0 m
Angular frequency, ω = 400 rad s⁻¹
Magnetic field, B = 0.5 T (parallel to the axis of rotation)

Consider a small element of the rod of length dr at a distance r from the axis of rotation. The velocity of this element is v = rω. Since the magnetic field is parallel to the axis of rotation, it is perpendicular to the plane of rotation of the rod and thus perpendicular to the velocity of each element of the rod. Therefore, the motional emf induced across this small element is dE = Bv dr = B(rω) dr.

To find the total emf developed between the centre (r=0) and the ring (r=L), we integrate dE from 0 to L.

Question 13

Q13. A circular coil of 100 turns and radius 10 cm is placed with its plane normal to the magnetic field of 0.5 T. If the magnetic field is reversed in direction in 0.2 s, calculate the

Accepted Answer

The emf induced in a coil is given by Faraday's Law of Electromagnetic Induction, which states that the magnitude of the induced emf is equal to the rate of change of magnetic flux through the coil.

Given:
Number of turns, N = 100
Radius of the coil, r = 10 cm = 0.1 m
Magnetic field magnitude, B = 0.5 T
Time taken for reversal, Δt = 0.2 s

Initial state: The plane of the coil is normal to the magnetic field. This means the area vector of the coil is parallel to the magnetic field. So, the angle θ between the magnetic field and the area vector is 0°.
Final state: The magnetic field is reversed. This means the angle θ between the magnetic field and the area vector becomes 180°.

Question 14

Q14. A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direct

Accepted Answer

The induced emf in the loop is 1.0 × 10⁻³ V, and the induced current is 2.0 × 10⁻³ A.

Question 15

Q15. A circular coil of radius 10 cm, 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the Earth’s magnetic field. It is rotated about

Accepted Answer

The magnitude of the induced emf is 3.77 × 10⁻³ V, and the magnitude of the induced current is 1.88 × 10⁻³ A.

Question 16

Q16. A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direct

Accepted Answer

The magnitude of the induced emf is approximately 1.01 mV, and the magnitude of the induced current is approximately 2.02 mA.

Question 17

Q17. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s⁻¹, at right angles to the horizontal component of the earth's magnetic field, 0.

Accepted Answer

(a) The instantaneous value of the induced emf is 1.5 mV.
(b) The direction of the induced emf (and current) is from East to West.
(c) The East end of the wire is at the higher electrical potential.

Chapter 6: Electromagnetic Induction