Question

A circular loop of wire with a radius of 4.0 cm is in a uniform magnetic field of magnitude 0.069 t. the plane of the loop is perpendicular to the direction of the magnetic field. in a time interval of 0.46 s, the magnetic field changes to the opposite direction with a magnitude of 0.044 t. what is the magnitude of the average emf induced in the loop?

Answer 1

The magnitude of the average emf induced in the loop is given by (we ignore the signs since we are interested only in the magnitude)

`\epsilon = (\Delta \Phi_B)/(\Delta t)`
where
`\Delta \Phi_B` is the variation of magnetic flux through the area enclosed by the loop, and
`\Delta t` is the time interval.

The magnetic flux is given by

`\Phi _B = BA\cos \alpha`
where B is the intensity of the magnetic field, A is the area enclosed by the loop and
`\alpha` is the angle between the perpendicular to the area and the magnetic field. In our problem, this angle is zero because the loop is perpendicular to the magnetic field, so the cosine is 1. The area of the loop is fixed, and it is

`A=\pi r^2`
where
`r=4.0 cm=0.04 m` is the radius of the loop. The only element which is variable in the formula is B, which changes from 0.069 T to -0.044 T (opposite direction). So we can rewrite the flux variation as

`\Delta \Phi_B = A \Delta B`
where
`\Delta B = 0.069 T-(-0.044 T)=0.113 T`
By using
`\Delta t=0.46 s`, we can find the magnitude of the emf induced:

`\epsilon = (A \Delta B)/(\Delta t)= ((\pi (0.04 m)^2)(0.113 T))/(0.46 s)=1.2 \cdot 10^(-3)V`