Question

The self-inductance of a coil having 400 turns is 10mH. The magnetic flux through the cross section of the coil corresponding to 2mA is
(A) $4 \times {10^{ - 5}}Wb$
(B) $2 \times {10^{ - 3}}Wb$
(C) $3 \times {10^{ - 5}}Wb$
(D) $8 \times {10^{ - 3}}Wb$

Answer 1

To find out the magnetic flux through the cross section of the coil, we need to know about the relation between the self-inductance and the magnetic flux. When there is a change in the current flowing through the solenoid, an emf is induced. This induced emf is the reason for self-inductance of the coil. Using the self-inductance of the solenoid, we can find the flux through the coil.
Formula used: The magnetic flux, $\phi = NLI$ where N is the number of turns of the coil, L is the inductance in Henry and I is the current passing through the coil.

Complete step by step answer:
The relation between the magnetic flux and the self-inductance of the coil is given by,
$\phi = NLI$
Now, given that,
Number of turns $N = 400$
Self-inductance $L = 10mH = 10 \times {10^{ - 3}}H$
Current, $I = 2mA = 2 \times {10^{ - 3}}A$
So, the total magnetic flux is given by,
$\phi = 400\left( {10 \times {{10}^{ - 3}}} \right) \times 2 \times {10^{ - 3}}$
$\Rightarrow \phi = 8 \times {10^{ - 3}}Wb$
Hence, the correct answer is option (D).

Note:
The relation between the magnetic flux and the self-inductance should be known to solve this problem. The self-inductance concept is applied in various places like transformers, sensors, induction motors etc. The factors affecting self-inductance are the number of turns in the coil, the diameter of the coil, the coil length, the type of material used in the core, and the number of layers of winding in the coils.