A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of $8 c m s^{- 1}$ in the positive x- direction in an environment containg a magnetic field in the positive z- direction. The field is neither uniform in space nor constant in time. If has a gradient of $10^{- 3} T c m^{- 1}$ along the negative x -direction (that is it increases by $10^{- 3} T c m^{- 1}$ as one moves in the negative x -direction), and it is decreasing in time at the rate of $10^{- 3} T s^{- 1} .$ The magnitude of the induced current in the loop if its resistance is $4.5 m Ω$ is ------------------mA.

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answer is 29.

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Detailed Solution

$\begin{array}{l} ϕ = \int_{0}^{a} a B (x, t) d x \\ \frac{d ϕ}{d t} = a \int_{0}^{a} d x \frac{d B (x, t)}{d t} = a \int d x (\frac{\partial B (x, t)}{\partial t} + \frac{\partial B}{\partial x} \times \frac{d x}{d t}) \\ = A [\frac{\partial B}{\partial t} + V \frac{\partial B}{\partial x}] = 144 \times 10^{- 4} [10^{- 3} \times 8 + 10^{- 3}] \\ = 1296 \times 10^{- 7} \\ ∴ I = \frac{E}{R} = \frac{1296 \times 10^{- 7}}{4.5 \times 10^{- 3}} = 288 \times 10^{- 4} = 28.8 m A \end{array}$