SATHEE: Electro Magnetic Induction And Alternating Currents Question 464

Electro Magnetic Induction And Alternating Currents Question 464

Question: The magnetic field in a region is given by $B = B_{0} (1 + \frac{x}{a}) \hat{k}$ . A square loop of edge-length d is placed with its edges along the x and y-axes. The loop is moved with a constant velocity $v = v_{0} \hat{i} .$ The emf induced in the loop is:

Options:

A) zero

B) $v_{0} B_{0} d$

C) $\frac{v_{0} B_{0} d^{3}}{a^{2}}$

D) $\frac{v_{0} B_{0} d^{2}}{a}$

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Answer:

Correct Answer: D

Solution:

At $x = 0$ , $B = B_{0}$ , so $e_{1} = B_{0} v_{0} d$

At, $x = d$ , $B = B_{0} (1 + \frac{d}{a})$ , so $e_{2} = B_{0} (1 + \frac{d}{a}) v_{0} d$

Now $e_{n e t} = e_{2} - e_{1} = \frac{B_{0} v_{0} d^{2}}{a}$

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Electro Magnetic Induction And Alternating Currents Question 464

Question: The magnetic field in a region is given by B=B0(1+xa)k^ . A square loop of edge-length d is placed with its edges along the x and y-axes. The loop is moved with a constant velocity v=v0i^. The emf induced in the loop is:

Options:

Answer:

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Question: The magnetic field in a region is given by $B = B_{0} (1 + \frac{x}{a}) \hat{k}$ . A square loop of edge-length d is placed with its edges along the x and y-axes. The loop is moved with a constant velocity $v = v_{0} \hat{i} .$ The emf induced in the loop is: