A square loop of side length Land resistance R is moving with a constant velocity v in a region containing a magnetic field B = B0(x/L) k-hat, where x is the distance from the origin. The loop moves in the positive x-direction. Find the magnitude of the induced current in the loop when its leading edge is at position x = d.

A. B_0 v LR — The induced current is calculated by finding the rate of change of magnetic flux through the loop as it moves through the non-uniform field.

JEE Main · PhysicsHard

$A square loop of side length L and resistance R is moving with a constant velocity v in a region containing a magnetic field B = B_{0} \frac{x}{L} \hat{k}, where x is the distance from the origin. The loop moves in the positive x -direction. Find the magnitude of the induced current in the loop when its leading edge is at position x = d .$

A. $\frac{B _{0} v L}{R}$
B. $\frac{B _{0} v L ^{2}}{R}$
C. $\frac{B _{0} v d}{R}$
D. $\frac{B _{0} v L}{2 R}$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{B _{0} v L}{R}$

Solution

$The magnetic flux through the loop at position x = d is Φ = \int_{d - L}^{d} (B_{0} x / L) L d x .$
$Evaluating the integral gives Φ = \frac{B _{0} L}{L} [x^{2} /2]_{d - L}^{d} = B_{0} (d^{2} - (d - L)^{2}) /2 = B_{0} (2 d L - L^{2}) /2 = B_{0} L (d - L /2) .$
$The induced EMF is E = - \frac{d Φ}{d t} = - \frac{d Φ}{d x} \frac{d x}{d t} = - B_{0} Lv .$
$The magnitude of the current is I = ∣ E ∣/ R = B_{0} v L / R .$
$Hence the answer is (A).$

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