Question

A coil having resistance $40\Omega$, number of turns 100 and radius 6 mm is connected to an ammeter of resistance $160\Omega$. The coil is placed perpendicular to the magnetic field. When the coil is taken out of the field, a charge of $32\mu C$ passes through it. The intensity of magnetic field will be

• A. 6.55 T
• B. 5.66 T
• C. 0.655 T
• D. 0.566 T

Answer 1

Answer: (D)

0.566 T

$Given:$ Radius $r=6mm$

Resistance of coil $=40$ $\Omega$

Number of turns $=100$

Resistance of ammeter $=160$ $\Omega$

Total resistance $R=200$ $\Omega$

Charge $q=32\times {10}^{-6}C$

$Solution:$ We know the induced emf is

$\zeta =\frac{\Delta \varphi }{\Delta t}$

$IR=\frac{\Delta \left(BA\right)}{\Delta t}$

$\frac{\Delta Q}{\Delta t}R=\frac{\delta (B\cdot A)}{\Delta t}$

$\Delta B=\frac{\Delta Q\cdot R}{A}$

The area is

$A=N\pi r^2$

$A=100\times 3.14\times {(6\times {10}^{-3})}^2$

$A=0.011304$

The magnetic flux density is

$\Delta B=\frac{32\times {10}^{-6}\times 200}{0.011304}$

$\Delta B=0.566T$

Hence, The magnetic flux density is 0.566 T.

$So,the$ $correct$ $option:D$