Question

A single circular loop with a radius of 35 cm is placed in a uniform external magnetic field with a strength of 0.50 T so that the plane of the coil is perpendicular to the field. The coil is pulled steadily out of the field in 15 s. Find the magnitude of the average induced emf during this interval. Show all work and include units of measure.

Answer 1

0.0129 V

Step-by-step explanation:

The magnitude of the induced emf in the circuit is given by:

`\epsilon = (\Delta \Phi)/(\Delta t)`

where

`\Delta \Phi` is the change in magnetic flux through the coil

`\Delta t` is the time interval

To find the change in magnetic flux, we need to find the initial flux and the final flux.

The area of the coil is

`A=\pi r^2 = \pi (0.35 m)^2=0.385 m^2`

The initial magnetic field is

`B_i = 0.50 T`

so the initial flux is

`\Phi_i = B_i A = (0.50 T)(0.385 m^2)=0.193 Wb`

While the final flux is zero, since the coil is completely out of the magnetic field:

`\Phi_f = 0`

so the magnitude of the change in flux is

`\Delta \Phi = |\Phi_f - \Phi_i|=|0-0.193 Wb|=0.193 Wb`

While the time interval is

`\Delta t = 15 s`

so the induced emf is

`\epsilon = (0.193 Wb)/(15 s)=0.0129 V`