Question

A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.

Answer 1

Answer:

$0.2677\ \text{V/m}$

Explanation:

A = Area of loop = $0.129\times0.402$

B = Magnetic field = $0.888\ \text{T}$

t = Time taken = $0.172\ \text{s}$

Electric field is given by

$E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}$

The emf induced is $0.2677\ \text{V/m}$.