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)

$4\times {10}^{-5}$ Wb

Here,

$N=400,L=10mH=10\times {10}^{-3}H,I=2mA=2\times {10}^{-3}A$

total magnetic flux with the coil,
$\varphi =NLI=400\times (10\times {10}^{-3})\times 2\times {10}^{-3}=8\times {10}^{-3}Wb$
Magnetic flux through the cross - section of the coil
= Magnetic flux linked with each turn =$\frac{\varphi }{N}=\frac{8\times {10}^{-3}}{200}=4\times {10}^{-5}Wb$