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Physics Question on Electromagnetic induction

The pole strength of 12 cm12 \,cm long bar magnet is 20 A m20 \,A \,m. The magnetic induction at a point 10 cm10\, cm away from the centre of the magnet on its axial line is [μ04π=10−7 H m−1]\left[\frac{\mu_{0}}{4\pi} = 10^{-7}\,H\,m^{-1}\right]

A

1.17×10−3 T1.17 \times 10^{-3}\, T

B

2.20×10−3 T2.20 \times 10^{-3}\, T

C

1.17×10−2 T1.17 \times 10^{-2}\, T

D

2.20×10−2 T2.20 \times 10^{-2}\, T

Answer

1.17×10−3 T1.17 \times 10^{-3}\, T

Explanation

Solution

On axial line, B=μ04π2md(d2−l2)2B = \frac{\mu_{0}}{4\pi} \frac{2md}{\left(d^{2} - l^{2}\right)^{2}} Given, 2l=12 cm=0.12 m2l = 12\, cm = 0.12 \,m Pole strength =20 A m= 20 \,A \,m, d=10 cm=0.1 md = 10 \,cm = 0.1\, m Magnetic moment m=20×0.12 Am2m = 20 \times 0.12 \,Am^{2} ∴B=10−7×2(20)×(0.12)×0.1[(0.1)2−(0.06)2]2=1.17×10−3 T\therefore\quad B = 10^{-7} \times \frac{2\left(20\right) \times \left(0.12\right) \times 0.1}{\left[\left(0.1\right)^{2} - \left( 0.06\right)^{2}\right]^{2}} = 1.17 \times 10^{-3}\,T