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

Answer –

Length of the wired loop is given by l = 8 cm = 0.08 m

Width of the wired loop is given by b = 2 cm = 0.02 m

Since the loop is a rectangle, the area of the wired loop is given by –

A = lb = 0.08 × 0.02

A = 16 × 10-4 m2

Strength of magnetic field is given by B = 0.3 T

Velocity of the loop v is given to be = 1 cm / s = 0.01 m / s

(i) Therefore, Emf developed in the loop is given as by the relation –

e = Blv = 0.3 × 0.08 × 0.01

e=2.4\times {{10}^{-4}}V

Time taken to travel along the width is given by –

t=\frac{b}{v}=\frac{0.02}{0.01}=2s

Therefore, the induced voltage is 2.4×10-4 V which lasts for 2 s.

ii) Emf developed is given by –

e = Bbv

e = 0.3 × 0.02 × 0.01 = 0.6 × 10-4V

Time taken to travel along the length is given by the expression –

t=\frac{l}{v}=\frac{0.08}{0.01}=8s

Therefore, the induced voltage is 0.6×10−4V  which lasts for 8 s.