Question

A small flat search coils of area 4 cm2 with 20 closely wound turns is positioned normal to the field direction and then quickly snatched out of the field region. The total charge flowing in the coil (measured by a ballistic galvanometer connected to the coil) is 7.5 mC. The resistance of the coil and galvanometer is 0.8 $\Omega$The field strength of the magnet is

• A. 1.25 T
• B. 0.50 T
• C. 0.75 T
• D. 2.10 T

Answer 1

Answer: (C)

0.75 T

Answer 2

Here , $A = 4cm^{2} = 4 \times 10^{- 4}m^{2},N = 20$

Final flux $\varphi_{f} = 0$

(when the coil is removed from the field)

$q = 7.5mC = 7.5 \times 10^{- 3}C,R = 0.8\Omega$

As $I = \frac{\varepsilon}{R} = \frac{- N(d\varphi/dt)}{R}$

$Idt = - \frac{N}{R}d\varphi$

Also charge,

$q = \int_{}^{}{Idt = \int_{\varphi_{i}}^{\varphi_{f}}{- \frac{N}{R}d\varphi = - \frac{N}{R}(\varphi_{f} - \varphi_{i}) = \frac{N}{R}(\varphi_{i} - \varphi_{f})}}q = \frac{N}{R}\varphi_{i}$

Now, initial flux per turn when coil is normal to the

field

$\theta_{i} = BA$

$\therefore q = \frac{NBA}{R}$

Or $B = \frac{qR}{NA} = \frac{7.5 \times 10^{- 3} \times 0.8}{20 \times 4 \times 10^{- 4}} = 0.75T$