It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet.
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 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° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.
Area of the small flat search coil, A = 2 cm2 = 2 × 10−4 m2
Number of turns on the coil, N = 25
Total charge flowing in the coil, Q = 7.5 mC = 7.5 × 10−3 C
Total resistance of the coil and galvanometer, R = 0.50 Ω
Induced current in the coil,
$I=\frac{\text { Induced emf }(e)}{R}$ ...(i)
Induced emf is given as:
$e=-N \frac{d \phi}{d t}$ ...(ii)
Where,
$d \phi=$ Charge in flux
Combining equations (1) and (2), we get
$I=-\frac{N \frac{d \phi}{d t}}{R}$
$I d t=-\frac{N}{R} d \phi$ ...(iii)
Initial flux through the coil, $\phi_{i}=B A$
Where,
B = Magnetic field strength
Final flux through the coil, $\phi_{f}=0$
Integrating equation (3) on both sides, we have
$\int I d t=\frac{-N}{R} \int_{\phi}^{\phi_{f}} d \phi$
But total charge, $Q=\int I d t$.
$\therefore Q=\frac{-N}{R}\left(\phi_{f}-\phi_{i}\right)=\frac{-N}{R}\left(-\phi_{i}\right)=+\frac{N \phi_{i}}{R}$
$Q=\frac{N B A}{R}$
$\therefore B=\frac{Q R}{N A}$
$=\frac{7.5 \times 10^{-3} \times 0.5}{25 \times 2 \times 10^{-4}}=0.75 \mathrm{~T}$
Hence, the field strength of the magnet is 0.75 T.