## Electromagnetic Induction class 12 ncert solutions: Ncert Class 12 Physics Chapter 6 Exercise Solutions

Textbook | NCERT |

Class | Class 12 |

Subject | Physics |

Chapter | Chapter 6 |

Chapter Name | Electromagnetic Induction class 12 exercise solutions |

Category | Ncert Solutions |

Medium | English |

Are you looking for Ncert Solution for Class 12 Physics Chapter 6 Electromagnetic Induction? Now you can download Ncert Class 12 Physics Chapter 6 Exercise Solutions pdf from here.

### Question 6.1: Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f ).

**Solution 6.1:** Lenz’s law shows the direction of the induced current in a closed loop. The given two figures show the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed loop, respectively.

We can predict the direction of induced current in different situations by using Lenz’s rule:

- (i) The direction of the induced current is along qrpq.
- (ii) The direction of the induced current is along prq along yzx.
- (iii) The direction of the induced current is along yzxy.
- (iv) The direction of the induced current is along zyxz.
- (v) The direction of the induced current is along xryx.
- (vi) No current is induced since the field lines are lying in the plane of the closed loop.

### Question 6.2: Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.

**Solution 6.2:** According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

(a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.

(b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along adcb.

### Question 6.3: A long solenoid with 15 turns per cm has a small loop of area 2.0 cm^{2 }placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

**Solution 6.3:**

- Number of turns on the solenoid – 15 turns/cm = 1500 turns / m
- Number of turns per unit length, n = 1500 turns
- The solenoid has a small loop of area, A = 2.0 cm
^{2}= 2 × 10^{−4}m^{2} - The current carried by the solenoid changes from 2 A to 4 A.
- Therefore, Change in current in the solenoid, di = 4 – 2 = 2 A
- Change in time, dt = 0.1 s

According to Faraday’s law, induced emf in the solenoid is given by:

\(e = \frac{d\phi }{dt} \;\;\;\;\;\;\;\;\; . . . (1)\)

Where, \(\phi\) = Induced flux through the small loop

= BA . . . (2)

B = Magnetic field

= \(\mu _{0} ni\)

μ_{0} = Permeability of free space

= \(4\pi \times 10 ^{-7} \; H/m\)

Hence, equation (1) can be reduced to: \(e = \frac{d}{dt}\left ( BA \right )\)

\(e = A\, \mu _{0}\, n \times \left (\frac{di}{dt} \right ) \ \ \)

\(= 2 \times 10^{-4} \times 4 \pi \times 10 ^{-7} \times 1500 \times \frac{2}{0.1}\ \ \)

= 7.54 **×** 10^{-6} V

Hence, the induced voltage in the loop is 7.54 **×** 10^{-6} V.

### Question 6.4: A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed, normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^{–1} in a direction normal to the (a) longer side and (b) shorter side of the loop? For how long does the induced voltage last in each case?

**Solution 6.4:**

- Length of the wired loop, l = 8 cm = 0.08 m
- Width of the wired loop, b = 2 cm = 0.02 m
- Since the loop is a rectangle, the area of the wired loop,
- A = lb
- = 0.08 × 0.02
- = 16
**×**10^{-4}m^{2}

Strength of magnetic field, B = 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m/s

**(a) **Emf developed in the loop is given as:

- e = Blv
- = 0.3 × 0.08 × 0.01
- = 2.4
**×**10^{-4}V

Time taken to travel along the width ,

\(t = \frac{Distance \; travelled}{Velocity} = \frac{b}{v} \ \ \)

\(= \frac{0.02}{0.01} = 2s\)

Hence, the induced voltage is 2.4 **×** 10^{-4} V, which lasts for 2 s.

**(b) **Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01

= 0.6 **×** 10^{-4} V

The time taken to travel along the length,

\(t = \frac{Distance \; travelled}{Velocity} = \frac{l}{v} \ \ \)

\(= \frac{0.08}{0.01} = 8s\)

Hence, the induced voltage is 0.6 **×** 10^{-4} V, which lasts for 8 s.

### Question 6.5: 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 with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

**Solution 6.5:** To calculate the emf developed between the center and the ring, we can use the concept of motional emf generated due to the motion of a conductor in a magnetic field.

**Given:**

- Length of the rod, L = 1.0 m
- Angular frequency, ω = 400 rad/s
- Magnetic field strength, B = 0.5 T

**Concept:**

When a rod rotates in a magnetic field, an emf is induced across its length due to the magnetic flux cutting the conductor. The emf induced between the center of the rod and the ring can be calculated using the formula:

emf = \(\frac{1}{2} B \omega L^2\)

**Explanation:**

- The magnetic field ( B ) is parallel to the axis of rotation.
- The emf generated in a small element of the rod of length ( dx ) at a distance ( x ) from the axis is demf = Bv( x ) dx, where v( x ) = ωx is the linear velocity of the element.
- Integrating from ( x = 0 ) (the center) to ( x = L ) (the end of the rod) gives the total emf between the ends.

**Calculation:**

Substitute the given values into the formula:

emf = \(\frac{1}{2} \times 0.5 \, \text{T} \times 400 \, \text{rad/s} \times (1.0 \, \text{m})^2\)

emf = \( \frac{1}{2} \times 0.5 \times 400 \times 1.0^2\)

emf = \(\frac{1}{2} \times 200 \, \text{V}\)

emf = 100 V

The emf developed between the center and the ring is **100 V**.

### Question 6.6: 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, 0.30 × 10^{–4} Wb m^{–2}.

(a) What is the instantaneous value of the emf induced in the wire?

(b) What is the direction of the emf?

(c) Which end of the wire is at the higher electrical potential?

**Solution 6.6:** Wire’s length, l = 10 m

Speed of the wire with which it is falling, v = 5.0 m/s

Strength of magnetic field, B = 0.3 **×** 10^{-4}Wb m^{–2}

**(a)** EMF induced in the wire,

- e = Blv
- = 0.3
**×**10^{-4}**×**5**×**10 - = 1.5
**×**10^{-3}v

(b) We can determine the direction of the induced current by using Fleming’s right-hand thumb rule; here, the current is flowing in the direction from West to East.

(c) In this case, the eastern end of the wire will have a higher potential.

### Question 6.7: Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

**Solution 6.7:** To find the self-inductance ( L ) of the circuit, we can use Faraday’s Law of Electromagnetic Induction, which relates the induced emf ( *ϵ* ) to the rate of change of current through the circuit.

**Given:**

- Initial current, \( I_1 = 5.0 \, \text{A} \)
- Final current, \( I_2 = 0.0 \, \text{A} \)
- Time interval, \( \Delta t = 0.1 \, \text{s} \)
- Induced emf,
*ϵ*= 200 V

**Faraday’s Law:**

The induced emf in a circuit is given by:

\(\epsilon = -L \frac{dI}{dt}\)

Here, \(\frac{dI}{dt}\) is the rate of change of current.

**Calculation:**

First, calculate the rate of change of current:

\(\frac{dI}{dt} = \frac{I_2 – I_1}{\Delta t} \)

\(= \frac{0.0 \, \text{A} – 5.0 \, \text{A}}{0.1 \, \text{s}} \)

\(= \frac{-5.0 \, \text{A}}{0.1 \, \text{s}} \)

= -50 A/s

Now, using Faraday’s Law:

\(\epsilon = L \times \left(- \frac{dI}{dt}\right)\)

Substitute the values:

200 V = L **×** 50 A/s

\(L = \frac{200 \, \text{V}}{50 \, \text{A/s}} = 4 \, \text{H}\)

The self-inductance of the circuit is **4 H**.

### Question 6.8: 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?

**Solution 6.8:** The change in flux linkage with the other coil due to the change in current in the first coil can be calculated using the formula for mutual inductance.

**Given:**

- Mutual inductance, \(M = 1.5 \, \text{H}\)
- Initial current in the first coil, \( I_1 = 0 \, \text{A}\)
- Final current in the first coil, \(I_2 = 20 \, \text{A}\)
- Time interval, \(\Delta t = 0.5 \, \text{s}\)

**Formula:**

The change in flux linkage \(\Delta \Phi\) with the second coil is given by:

\(\Delta \Phi = M \Delta I\)

where \(\Delta I\) is the change in current in the first coil.

**Calculation:**

First, calculate the change in current:

\(\Delta I = I_2 – I_1 = 20 \, \text{A} – 0 \, \text{A} = 20 \, \text{A}\)

Now, using the formula:

\(\Delta \Phi = M \times \Delta I = 1.5 \, \text{H} \times 20 \, \text{A}\)

\(\Delta \Phi = 30 \, \text{Wb-turns}\)

The change in flux linkage with the other coil is **30 Wb-turns**.