Question

The self inductance of an inductor coil having $100$ turns is $20\, mH$. The magnetic flux through the cross-section of the coil corresponding to a current of $4 \,mA$ is

• A. $2\times 10^{-5}\,Wb$
• B. $4\times 10^{-7}\,Wb$
• C. $8\times 10^{-7}\,Wb$
• D. $8\times 10^{-5}\,Wb$

Answer 1

Answer: (C)

Total magnetic flux,
$N\phi=LI=20\times10^{-3}\times4\times10^{-3}$
$=8\times10^{-5}\,Wb$
Magnetic flux through the cross-section of the coil,
$\phi=\frac{8\times10^{-5}}{N}=\frac{8\times10^{-5}}{100}$
$=8\times10^{-7}\,Wb$