Ncert Exercises & Solutions

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

Length of the wired loop, l = 8 cm = 0.08 m

Width of the wired loop, b = 2 cm = 0.02 m

Hence, the area of the röctangular loop, $A = l b = 0.08 \times 0.02 = 16 \times 10^{- 4} m^{2}$

Strength of magnetic field, B = 0.2 T

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

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

\begin{array}{l} e = Blv = 0 .3 \times 0 .08 \times 0 .01 \\ = 2 .4 \times 10^{- 4} V \end{array}

$\begin{array}{l} Time taken to travel along the width, t = \frac{Distance travelled}{Velocity} = \frac{b}{v} \\ = \frac{0 .02}{0 .01} = 2 s \end{array}$

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

(b) Emf developed,

e = Bbv = 0 .3 \times 0 .02 \times 0 .01 = 0 .6 \times 10^{- 4} V

\begin{array}{l} Time taken to travel along the length, t = \frac{Distance traveled}{Velocity} = \frac{l}{v} \\ = \frac{0 .08}{0 .01} = 8 s \end{array}

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

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