Question

A conducting coil of 1785 turns is connected to a galvanometer, and the total resistance of the circuit is 43.9 Ω. The area of each turn is 4.50 10-4 m2. This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced to flow around the circuit is measured to be 9.33 10-3 C. Find the magnitude of the magnetic field.

Answer 1

0.5099 T

Step-by-step explanation:

We are given that

Number of turns=N=1785

Resistance of circuit, R=
`43.9\Omega`

Area of each turn,
`A=4.5* 10^(-4) m^2`

Charge , q=
`9.33* 10^(-3) C`

We have to find the magnitude of the magnetic field.

We know that magnetic field,
`B=(qR)/(NA)`

Substitute the values

Magnetic field, B=
`(9.33* 10^(-3)* 43.9)/(1785* 4.5* 10^(-4))=0.5099 T`

Hence, the magnitude of the magnetic field=0.5099 T