A closely wound search coil (see Exercise 29.3) has an area of 3.20$\mathit{c}{\mathit{m}}^{\mathbf{2}}$, 120 turns, and a resistance of 60.0 Ω. It is connected to a charge-measuring instrument whose resistance is 45.0 Ω. When the coil is rotated quickly from a position parallel to a uniform magnetic field to a position perpendicular to the field, the instrument indicates a charge of$\mathbf{3}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}\mathbf{}}\mathit{C}$. What is the magnitude of the field?

Magnetic flux is expressed as;

$\mathrm{\Phi }=\mathrm{BAcos}\left(\mathrm{\Phi }\right)$

So, here B is the magnetic field

A is area.

Consider a search coil, which is used to determine the strength of a magnetic field. The coil's plane is initially held perpendicular to a magnetic field. The coil is then quickly rotated a quarter turn around a diameter or quickly pulled out of the field, reducing the flux through the coil from its maximum value to zero in a short period of time.

$$\begin{gathered} {\Phi }_{B,i}=BA\cos \left(\varphi \right)=BA \\ {\Phi }_{B,f}=0 \end{gathered}$$

The magnitude of the average emf induced in the coil equals the flux change divided by the time required to go from the initial flux to the final flux and multiplied by the number of turns N;

$\left|\epsilon av\right|=N\frac{\left|{\Phi }_{Bf}-{\Phi }_{Bi}\right|}{∆t}=\frac{NBA}{∆t}$

The induced current equals the induced emf divided by the resistance of the search coil, that is,

$l=\frac{\left|\epsilon =av\right|}{R}=\frac{NBA}{R∆t}$

So, the total charge flow is;

$$\begin{gathered} Q=l∆t=\left(\frac{NBA}{R∆t}\right)∆t=\frac{NBA}{R} \\ Q=\frac{NBA}{R} \end{gathered}$$

Hence, the total charge flow is_{$\frac{NBA}{R}$}

Coil area$A=3.20c{m}^2$

Number of turns N=120

Resistance is $60\Omega$and as it connected to a charge-measuring instrument whose resistance is$45\Omega$

Thus, the total resistance is R=105$\Omega$

On putting the values,

_{$$\begin{gathered} B=\frac{QR}{NA}=\frac{\left(3.56\times {10}^{-5}C\right)\left(105.0\Omega \right)}{120\left(3.20\times {10}^{-4}{m}^2\right)} \\ B=0.0937T \end{gathered}$$}

Hence, the magnitude of the field is B=0.0973T