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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Electromagnetic Induction

BITSAT 2015 · Physics · Electromagnetic Induction

A conducting square loop is placed in a magnetic field B with its...

A conducting square loop is placed in a magnetic field BB with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate α\alpha. The induced emf in the loop at an instant when its side is ' aa ' is

Options

  1. A2aαB2 a \alpha B
  2. Ba2αBa^{2} \alpha B
  3. C2a2αB2 a^{2} \alpha B
  4. DaαBa \alpha B

Correct answer

A. 2aαB2 a \alpha B

Step-by-step solution

At any time tt, the side of the square a=(a0−αt),a=\left(a_{0}\right.-\alpha t), where a0=a_{0}= side at t=0t=0 At this instant, flux through the square : ϕ=BAcos⁡0∘=B(a0−\alphat)2∴ emf induced E=−dϕdt⇒E=−B.2(a0−\alphat)(0−α)=+2\alphaaB\begin{array}{l} \phi=B A \cos 0^{{}^{\circ}}=B\left(a_{0}-\alphat\right)^{2} \\ \therefore \text{ emf induced }E=-\frac{d \phi}{d t}\\ \Rightarrow E=-B .2\left(a_{0}-\alphat\right)(0-\alpha)=+2 \alphaa B \end{array}
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