Question

A closed flat loop conductor with radius 2mm is located in a changing uniform magnetic field. If the emf induced in the loop is 2 V what is the rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies

Answer 1

Rate of magnetic field is 15923.668 T/sec

Step-by-step explanation:

We have given radius of loop conductor r = 2 mm = 0.002 m

Induced emf = 2 volt

As the magnetic field and plane are perpendicular to each other so angle between magnetic field and area is 0°

Cross sectional area of the conductor is equal to
`A=\pi r^2`

`A=3.14* 0.002^2=1.256* 10^(-6)m^2`

Induced emf is given by
`e=(-d\Phi )/(dt)=-A(dB)/(dt)`

`2=1.256* 10^(-6)* (dB)/(dt)`

`(dB)/(dt)=159235.668T/sec`