Question

The magnetic field at a point inside a $2\text{mH}$ inductor coil becomes $0.8$ of its maximum value in $20\mu \text{s}$ when the inductor is joined to a battery. Find the resistance of the circuit.
[Take $\ln \left(0.2\right)=-1.609$]

• A. $144\Omega$
• B. $127\Omega$
• C. $133\Omega$
• D. $161\Omega$

Answer 1

The correct option is D $161\Omega$

Let the resistance of the circuit is $R$ and the time constant of the circuit is $\tau$.

At any time, $t$ current through the $L-R$ circuit is,

$i=i_0(1-e^{-t/\tau }).......\left(1\right)$

As we know, the magnetic field inside a long coil (solenoid) is,

$B={\mu }_{0}ni...........\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get,

$B={\mu }_{0}n{i}_0(1-e^{-t/\tau })$

Assuming, $B_0={\mu }_{0}n{i}_0$

$B=B_0(1-e^{-t/\tau })........\left(3\right)$

Here,
$B=0.8B_0;L=2\text{mH};t=20\mu \text{s}$

Putting this values in $\left(3\right)$ we get,

$0.8B_0=B_0(1-e^{-20\times {10}^{-6}/\tau })$

$e^{-20\times {10}^{-6}/\tau }=0.2$

Taking $\log$ on both sides we get,

$-\frac{20\times {10}^{-6}}{\tau }=\ln \left(0.2\right)$

$\tau =\frac{-20\times {10}^{-6}}{-1.609}[\because \ln (0.2)=-1.609]$

The time constant of $L-R$ circuit is,

$\tau =\frac{L}{R}=\frac{-20\times {10}^{-6}}{-1.609}$

$\frac{2\times {10}^{-3}}{R}=\frac{20\times {10}^{-6}}{1.609}$

$\therefore R=160.9\Omega \approx 161\Omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.