A short bar magnet produces a magnetic field of $6.4 \times 10^{-5} \mathrm{~T}$ at a distance of $20…

1. $4.8 \times 10^{-5} \mathrm{~T}$
2. $3.2 \times 10^{-5} \mathrm{~T}$
3. $1.6 \times 10^{-5} \mathrm{~T}$
4. $6.4 \times 10^{-5} \mathrm{~T}$

Solution

Magnetic field produced by a magnet on its axis and its equator are given by $B_{\text {axis }}=\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{r_1^3}$ $B_{\text {equator }}=\frac{\mu_0}{4 \pi}=\frac{M}{r_2^3}$ $B_{\text {axis }}=B_{\text {equator }} \times \frac{2 r_2^3}{r_1^3}$ ...(i) Where, $M$ = magnetic dipole moment of magnet and $r=$ distance from centre of magnet
So, $\frac{B_{\text {axis }}}{B_{\text {equator }}}=\frac{\frac{\mu_0}{4 \pi} \frac{2 M}{r_1^3}}{\frac{\mu_0}{4 \pi} \cdot \frac{M}{r_2^3}}=\frac{2 r_2^3}{r_1^3}$ $B_{\text {axis }}=B_{\text {equator }} \times \frac{2 r_2^3}{r_1^3}$...(ii) Here, given $\quad B_{\text {equator }}=6.4 \times 10^{-5} \mathrm{~T}$ $\begin{aligned} & r_1=40 \mathrm{~cm}=40 \times 10^{-2} \mathrm{~m} \\ & r_2=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\end{aligned}$ So from eq. (i), we have, $B_{\text {axis }}=\frac{6.4 \times 10^{-5} \times 2 \times\left(20 \times 10^{-2}\right)^3}{\left(40 \times 10^{-2}\right)^3}$ $=\frac{6.4 \times 10^{-5} \times 2 \times 2^3}{4^3}=1.6 \times 10^{-5} \mathrm{~T}$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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