A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4mm and a total length of 30 cm. The magnetic field changes with time at a steady rate $\frac{d B}{d t} = 0.032 T s^{- 1}$ . The induced current in the loop is close to
(Resistivity of the metal wire is $1.23 \times 10^{- 8} Ω m$ )

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0.34A

0.53A

0.43A

0.61A

answer is C.

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Detailed Solution

$\begin{array}{l} F l u x = ϕ = B ℓ^{2} \cos θ [F i e l d i s p e r p e n d i c u l a r t o t h e l o o p . S o, θ = 0^{0}] \\ I n d u c e d e m f = ε = \frac{d B}{dt} ℓ^{2} [h e r e ℓ = s i d e o f s q u a r e] \\ I n d u c e d c u r r e n t = i = \frac{ε}{R} \\ \Rightarrow i = \frac{dB}{dt} \frac{ℓ^{2}}{\frac{(ρ L)}{A}} [h e r e L = t o t a l p e r i m e t e r o f s q u a r e, ℓ = \frac{L}{4}] \\ \Rightarrow i = \frac{dB}{dt} \frac{LA}{16 ρ} \\ \Rightarrow i = \frac{0.032 \times 0.3 \times π \times 2^{2} \times 10^{- 6}}{16 \times 1.23 \times 10^{- 8}} = 0.61 A \end{array}$