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?
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 $8 \mathrm{~s}$.
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