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

`0.2677\ \text{V/m}`

Step-by-step explanation:

A = Area of loop =
`0.129*0.402`

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

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

Electric field is given by

`E=B(dA)/(dt)\\\Rightarrow E=0.888*(0.129* 0.402)/(0.172)\\\Rightarrow E=0.2677\ \text{V/m}`

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