Solveeit Logo
Login

Question

Question: The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section...

The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA is

A

4 x 10-5 Wb

B

2 x 10-3 Wb

C

3 x 10-5 Wb

D

8 x 10-3 Wb

Answer

4 x 10-5 Wb

Explanation

Solution

Here, N = 400, L = 10 mH =10×10−3H= 10 \times 10 ^ { - 3 } \mathrm { H }

I=2 mA=2×10−3 A\mathrm { I } = 2 \mathrm {~mA} = 2 \times 10 ^ { - 3 } \mathrm {~A}

Total magnetic flux linked with the coil.

ϕ=NLI=400×(10×10−3)×2×10−3=8×10−3 Wb\phi = \mathrm { NLI } = 400 \times \left( 10 \times 10 ^ { - 3 } \right) \times 2 \times 10 ^ { - 3 } = 8 \times 10 ^ { - 3 } \mathrm {~Wb}magnetic flux through the cross section of the coil.

= Magnetic flux linked with each turn

=ϕN=8×10−3200=4×10−5 Wb= \frac { \phi } { \mathrm { N } } = \frac { 8 \times 10 ^ { - 3 } } { 200 } = 4 \times 10 ^ { - 5 } \mathrm {~Wb}