Page No 232: - Chapter 6 Electromagnetic Induction Additional Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 6.15:

An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10⁻³ s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Answer:

Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 cm² = 25 × 10⁻⁴ m²

Number of turns on the solenoid, N = 500

Current in the solenoid, I = 2.5 A

Current flows for time, t = 10⁻³ s

Average back emf, 

Where,

 = Change in flux

= NAB … (2)

Where,

B = Magnetic field strength

Where,

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

Using equations (2) and (3) in equation (1), we get

Hence, the average back emf induced in the solenoid is 6.5 V.

Question 6.16:

(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.

Calculate the induced emf in the loop at the instant when x = 0.2 m.

Take a = 0.1 m and assume that the loop has a large resistance.

Answer:

(a) Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).

Magnetic flux associated with element

Where,

dA = Area of element dy = a dy

B = Magnetic field at distance y

I = Current in the wire

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

y tends from x to .

(b) Emf induced in the loop, e = B’av

Given,

I = 50 A

x = 0.2 m

a = 0.1 m

v = 10 m/s

Question 6.17:

A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = − B₀ k (r ≤ a; a < R)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

Answer:

Line charge per unit length 

Where,

r = Distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, 

At distance r, the magnetic force is balanced by the centripetal force i.e.,

∴Angular velocity,