Question:

A rectangular wire loop of sides $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ 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 developed across the cut if the velocity of the loop is 1 $\mathrm{cm} \mathrm{s}^{-1}$ in a direction normal to the

(a) longer side,

(b) shorter side of the loop? For how long does the induced voltage last in each case?

Solution:

Length of the rectangular wire, l = 8 cm = 0.08 m

Width of the rectangular wire, b = 2 cm = 0.02 m

Hence, area of the rectangular loop,

A = lb

= 0.08 × 0.02

= 16 × 10−4 m2

Magnetic field strength, 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, $t=\frac{\text { Distance travelled }}{\text { Velocity }}=\frac{b}{v}$

$=\frac{0.02}{0.01}=2 \mathrm{~s}$

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

Time taken to travel along the length, $t=\frac{\text { Distance traveled }}{\text { Velocity }}=\frac{l}{v}$

$=\frac{0.08}{0.01}=8 \mathrm{~s}$

Hence, the induced voltage is $0.6 \times 10^{-4} \mathrm{~V}$ which lasts for $8 \mathrm{~s}$.