Question

A magnetic flux of $5$ microweber is linked with a coil when a current of $1\,mA$ flows through it. Calculate self-inductance of the coil.
(A) $5\,mH$
(B) $10\,mH$
(C) $15\,mH$
(D) $20\,mH$

Answer 1

Hint: The self-inductance is directly proportional to the magnetic flux and inversely proportional to the current. By using these two statements, the self-inductance can be determined. By the given terms in the question, these two parameters are used to determine the self-inductance of the coil.

Formulae Used:
The self-inductance of the coil is,
$L = \dfrac{\phi }{I}$
Where, $L$ is the self-inductance, $\phi $ is the magnetic flux and $I$ is the current.

 Complete step-by-step solution:
Given that,
The magnetic flux, $\phi = 5\,\mu Wb$
The current is flows in the coil is, $I = 1\,mA$
The self-inductance of the coil is,
$L = \dfrac{\phi }{I}\,.................\left( 1 \right)$
On substituting the magnetic flux and the current in the above equation (1), then the equation (1) is written as,
$L = \dfrac{{5\,\mu Wb}}{{1\,mA}}$
Here, the value of $\mu $ is ${10^{ - 6}}$ and the value of $m$ is ${10^{ - 3}}$, so substitute these values in the above equation, then the above equation is written as,
$L = \dfrac{{5 \times {{10}^{ - 6}}}}{{1 \times {{10}^{ - 3}}}}$
By taking the term ${10^{ - 3}}$ from denominator to the numerator for the further calculation so the sign of power gets changed, then the above equation is written as,
$L = \dfrac{{5 \times {{10}^{ - 6}} \times {{10}^3}}}{1}$
On multiplying the terms in the above equation, then the above equation is written as,
$L = \dfrac{{5 \times {{10}^{ - 3}}}}{1}$
On dividing the terms in the above equation, then the above equation is written as,
$L = 5 \times {10^{ - 3}}$
Then, the above equation is written as,
$L = 5\,mH$
Thus, the above equation shows the self-inductance of the coil.
Hence, the option (A) is the correct answer.

Note:- The self-inductance is directly proportional to the magnetic flux of the coil, so the magnetic flux increases the self-inductance also increases. Then, the self-inductance is inversely proportional to the current, so the current in the coil increases and the self-inductance decreases.