Answer to Question #276647 in Electricity and Magnetism for Sophia

Question

A 50-turn coil has a resistance 2.5 Ω. If it is enclosed in an area of 25 cm2, what would be the rate of change of a magnetic field parallel to its axis to induce a current of 0.5 A in the coil?

Answer 1

F=BA

where

F=rate of change of magnetic field

B= magnetic density

A= Area

$F=0.0025\times B$

The induced voltage per turn will be the time derivative of that value, so your total induced voltage will is

$V=50×0.0025×\frac{dB}{dt}$

From the formula V=IR we find that V=2.5×0.5=1.25v

Therefore:

$1.25=0.125\frac{dB}{dt}$

$\frac{dB}{dt}=\frac{1.25}{0.125}=10$

So rate of change of magnetic field= 10tesla/sec