Magnetic Properties of Matter

JEE Main level, single correct, medium

28 of 57

Question 28 of 57

A bar magnet of length 6 cm6 \mathrm{~cm} has a magnetic moment of 4 J T−14 \mathrm{~J} \mathrm{~T}^{-1}. Find the strength of magnetic field at a distance of 200 cm200 \mathrm{~cm} from the centre of the magnet along its equatorial line.

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Answer: (C) 5×10−85 \times 10^{-8} tesla

Along the equatorial line, magnetic field strength
B=μ04πM(r2+ℓ23/2)\left.B=\frac{\mu_0}{4 \pi} \frac{M}{\left(r^2+\ell^2{ }^{3 / 2}\right.}\right)
Given: M=4 JT−1M=4 \mathrm{~J} T^{-1}
r=200 cm=2 mr=200 \mathrm{~cm}=2 \mathrm{~m}
ℓ=6 cm2=3 cm=3×10−2 m\ell=\frac{6 \mathrm{~cm}}{2}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}
∴B=4π×10−74π×4[22+(3×10−2)2]3/2\therefore B=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{4}{\left[2^2+\left(3 \times 10^{-2}\right)^2\right]^{3 / 2}}
Solving we get, B=5×10−8B=5 \times 10^{-8} tesla

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