Question

A 50-turn coil with a radius of 15.1 cm is mounted so the coil’s axis can be oriented in any horizontal direction. Initially the axis is oriented so the magnetic flux from Earth’s field is maximized. If the coil’s axis is rotated through 90.0° in 0.080 s, an average emf of 0.547 mV is induced in the coil. What is the magnitude of the horizontal component of Earth’s magnetic field at this location?

Answer 1

Here,

N = 50

r = 15.1 cm = 0.151 m

t = 0.080 s

magnitude of magnetic field is B

emf = N * area * magnetic field * (cos(0) - cos(90))/time taken

0.547 *10^-3 = 50 * B * pi * 0.151^2 * 1/0.080

solving for B

B = 1.22 *10^-5 T

the magnitude of horizontal magnetic field is 1.22 *10^-5 T