It is desired to measure the magnitude of the magnetic field between the poles of a powerful loudspeaker magnet. $A$ small flat search coil of area $2 \; cm^{2}$ with $25$ closely wound turns is positioned normal to the field direction and then quickly snatched out of the field region. (Equivalently,one can give it a quick $90^{\circ}$ turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to the coil) is $7.5 \; mC$. The combined resistance of the coil and the galvanometer is $0.50 \; \Omega$. Estimate the field strength of the magnet.

(0.75 T) Given:
Area of the coil,$A = 2 \; cm^{2} = 2 \times 10^{-4} \; m^{2}$
Number of turns,$N = 25$
Total charge,$Q = 7.5 \; mC = 7.5 \times 10^{-3} \; C$
Total resistance,$R = 0.50 \; \Omega$
The induced emf is given by Faraday's law: $e = -N \frac{d\phi}{dt}$.
The induced current is $I = \frac{e}{R} = -\frac{N}{R} \frac{d\phi}{dt}$.
The total charge $Q$ is the integral of current over time: $Q = \int I \; dt = -\frac{N}{R} \int_{\phi_i}^{\phi_f} d\phi = -\frac{N}{R} (\phi_f - \phi_i)$.
Since the coil is removed from the field,the final flux $\phi_f = 0$ and the initial flux $\phi_i = BA$.
Thus,$Q = \frac{N \phi_i}{R} = \frac{NBA}{R}$.
Rearranging for the magnetic field $B$: $B = \frac{QR}{NA}$.
Substituting the values: $B = \frac{7.5 \times 10^{-3} \times 0.50}{25 \times 2 \times 10^{-4}} = \frac{3.75 \times 10^{-3}}{50 \times 10^{-4}} = \frac{3.75 \times 10^{-3}}{5 \times 10^{-3}} = 0.75 \; T$.
Therefore,the field strength of the magnet is $0.75 \; T$.