Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

# It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet.

Question:

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.

Solution:

Area of the small flat search coil, A = 2 cm= 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.