Question

An induced emf of 12V is produced when a 350 loop coil,

ofradius 4.2cm is placed perpendicular to a magnetic field and
turn90deg in 0.08s . calculate the intensity of the
megneticfield.

Answer 1

We can calculate the intensity of the magnetic field using this expression,

Induced EMF = nBπr²cos θ/Δt

where

n = number of turns of coil = 350

B = Magnetic field intensity =?

r = radius of the coil =4.2cm =0.042m

t = time taken for the change = 0.08s

θ = angle of the rotation in degree = 90deg

Now, note that the wire is perpendicular to the field initially, which gives the induced EMF, 12V but at angle 90deg, it gives another EMF =0V, because it is parallel here.

So, we will take an average of these two results to arrive at our answer.

12+0= (350*B*π*0.042²cos 0)/0.08 V

B = 12/24.255 = 0.495 Tesla

the intensity of the magnetic field = 0.495 Tesla