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

Correct answer: A. 2 × 10^ -5 ~Wb Here, N=400, L=10 mH=10 × 10^ -3 HI=2 mA=2 × 10^ -3 ATotal magnetic flux linked with the coil, aligned φ=N L I & =400 × (10 ×10^ -3 )×2 ×10^ -3 & =8 ×10^ -3 Wb aligned Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn =(φ)/(N)= 8 × 10^ -3 400 =2 ×10^ -5 Wb

AIIMS 2015

Exam: AIIMS
Year: 2015
Subject: Physics
Chapter: Electromagnetic Induction

AIIMS 2015 · Physics · Electromagnetic Induction

The self inductance of a coil having 400 turns is 10 mH . The...

The self inductance of a coil having 400 turns is

10 \mathrm{mH}

. The magnetic flux through the cross section of the coil corresponding to current

2 \mathrm{~mA}

Options

A $2 \times 10^{-5} \mathrm{~Wb}$
B $2 \times 10^{-3} \mathrm{~Wb}$
C $3 \times 10^{-5} \mathrm{~Wb}$
D $8 \times 10^{-3} \mathrm{~Wb}$

Correct answer

A. $2 \times 10^{-5} \mathrm{~Wb}$

Step-by-step solution

Here,

N=400, L=10 \mathrm{mH}=10 \times 10^{-3} \mathrm{H}I=2 \mathrm{ mA}=2 \times 10^{-3} \mathrm{ A}Total magnetic flux linked

with the coil,

\begin{aligned} \phi=N L I & =400 \times \left(10 \times10^{-3}\right)\times2 \times10^{-3} \\ & =8 \times10^{-3} \mathrm{ Wb} \end{aligned}

Magnetic flux through

the cross-section of the coil = Magnetic flux linked with each turn

=\frac{\phi}{N}=\frac{8 \times 10^{-3}}{400}=2 \times10^{-5} \mathrm{ Wb}

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