Question

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. $4\times 10^{-5}\,Wb$
• B. $2\times 10^{-3}\,Wb$
• C. $3\times 10^{-5}\,Wb$
• D. $8\times 10^{-3}\,Wb$

Answer 1

Answer: (A)

The relation between the magnetic flux and the self-inductance of the coil is given by,
$\phi=N L I$
Now, given that,
Number of turns $N=400$
Self-inductance $L=10 \,m H=10 \times 10^{-3} H$
Current, $I=2 \,m A=2 \times 10^{-3} A$
So, the total magnetic flux is given by,
$\phi=400\left(10 \times 10^{-3}\right) \times 2 \times 10^{-3}$
$\Rightarrow \phi=8 \times 10^{-3} W b$