Question

A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of 90°, so that the normal becomes perpendicular to the magnetic field. The coil has an area of 1.5 10-3 m2, 50 turns, and a resistance of 166 Ω. During the time while it is rotating, a charge of 7.3 10-5 C flows in the coil. What is the magnitude of the magnetic field?

Answer 1

Explanation:

Expression for magnitude of the induced emf is as follows.

             $\epsilon = N \frac{BA}{t}$

       $\frac{Q}{t}R = \frac{NBA}{t}$

So, magnitude of the magnetic field is as follows.

                   B = $\frac{RQ}{A \times N}$

It is given that,

       A = $1.5 \times 10^{-3} m^{2}$

       Q =$7.3 \times 10^{-5} C$

       N = 50

       R = 166 $\ohm$

Putting the given values into the above formula as follows.

              B = $\frac{RQ}{A \times N}$

                 = $\frac{166 \times 7.3 \times 10^{-5}}{1.5 \times10^{-3} \times 50}$

                = $\frac{1211.8 \times 10^{-5}}{75 \times 10^{-3}}$

                = $16.157 \times 10^{-2}$

                = 0.1615 T

Thus, we can conclude that magnitude of the magnetic field is 0.1615 T.