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Question: A magnetic flux of \(5\) microweber is linked with a coil when a current of \(1\,mA\) flows through ...

A magnetic flux of 55 microweber is linked with a coil when a current of 1 mA1\,mA flows through it. Calculate self-inductance of the coil.
(A) 5 mH5\,mH
(B) 10 mH10\,mH
(C) 15 mH15\,mH
(D) 20 mH20\,mH

Explanation

Solution

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=ϕIL = \dfrac{\phi }{I}
Where, LL is the self-inductance, ϕ\phi is the magnetic flux and II is the current.

Complete step-by-step solution :
Given that,
The magnetic flux, ϕ=5 μWb\phi = 5\,\mu Wb
The current is flows in the coil is, I=1 mAI = 1\,mA
The self-inductance of the coil is,
L=ϕI .................(1)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=5 μWb1 mAL = \dfrac{{5\,\mu Wb}}{{1\,mA}}
Here, the value of μ\mu is 10−6{10^{ - 6}} and the value of mm is 10−3{10^{ - 3}}, so substitute these values in the above equation, then the above equation is written as,
L=5×10−61×10−3L = \dfrac{{5 \times {{10}^{ - 6}}}}{{1 \times {{10}^{ - 3}}}}
By taking the term 10−3{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=5×10−6×1031L = \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=5×10−31L = \dfrac{{5 \times {{10}^{ - 3}}}}{1}
On dividing the terms in the above equation, then the above equation is written as,
L=5×10−3L = 5 \times {10^{ - 3}}
Then, the above equation is written as,
L=5 mHL = 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.