Question

When a coil of area $0.2{\text{cm}}^2$, having $30$ turns and whose plane is normal to the magnetic field is drawn out of the magnetic field, a charge of $1.5\times {10}^{-4}\text{C}$ flows in the circuit. If its resistance is $40\Omega$, then the magnetic flux density will be

• A. $10\text{T}$
• B. $0.1\text{T}$
• C. $1\text{T}$
• D. $0.01\text{T}$

Answer 1

The correct option is A $10\text{T}$

Given:

No. of turns, $N=30$

Area of coil, $A=0.2{\text{cm}}^2$

Charge, $q=1.5\times {10}^{-4}\text{C}$

Resistance of coil, $R=40\Omega$

The amount of charge flown through the loop,

$q=\frac{-\Delta \varphi }{R}$

Where, $\Delta \varphi ={\varphi }_f-{\varphi }_i$

As per the question final flux, ${\varphi }_f=0$

And initial flux, ${\varphi }_i=NBA$

$\therefore q=\frac{-({\varphi }_f-{\varphi }_i)}{R}$

$\Rightarrow 1.5\times {10}^{-4}=\frac{NBA}{40}$

$\Rightarrow B=\frac{6\times {10}^{-3}}{30\times 0.2\times {10}^{-4}}=10\text{T}$

Hence, option $\left(A\right)$ is correct.