Suppose the coil rotates from a position parallel to the uniform magnetic field to one perpendicular in time $\Delta t$. Since $\Delta t$ is very short, we have

$$
\varepsilon=\frac{\Delta \phi}{\Delta t}=i(R+r)
$$

As $q=i \Delta t$, the increase of the magnetic flux is

$$
\Delta \phi=q(R+r)=B A N,
$$

since the coil is now perpendicular to the field. Hence the magnetic flux density is

$$
\begin{aligned}
B &=\frac{(R+r) q}{A N}=\frac{(50+30) \times\left(4 \times 10^{-5}\right)}{4 \times 10^{-4} \times 160} \\
&=0.05 \mathrm{~T}=50 \mathrm{Gs} .
\end{aligned}
$$
A closely wound search coil has an area of $4 \mathrm{~cm}^{2}, 160$ turns and a resistance of $50 \Omega$. It is connected to a ballistic galvanometer whose resistance is $30 \Omega$. When the coil rotates quickly from a position parallel to a uniform magnetic field to one perpendicular, the galvanometer indicates a charge of $4 \times 10^{-5} \mathrm{C}$. What is the flux density of the magnetic field?