Question

The self induction 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. $2\times {10}^{-5}Wb$
• B. $8\times {10}^{-3}Wb$
• C. $4.3\times {10}^{-5}Wb$
• D. $4.8\times {10}^{-3}Wb$

Answer 1

Answer: (B)

$8\times {10}^{-3}Wb$

Given,

$N=400$

$L=10mH=10\times {10}^{-3}H$

$I=2mA=2\times {10}^{-3}A$

The magnetic flux through the coil,

$\varphi =NIL$

$\varphi =400\times 2\times {10}^{-3}\times 10\times {10}^{-3}$

$\varphi =8\times {10}^{-3}Wb$

The correct option is B.