JEE PYQ: Magnetism And Matter Question 1
Question 1 - 2021 (16 Mar 2021 Shift 1)
A bar magnet of length $14$ cm is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. A neutral point is obtained at a distance of $18$ cm from the center of the magnet. If $B_H = 0.4,\text{G}$, the magnetic moment of the magnet is $(1,\text{G} = 10^{-4},\text{T})$
(1) $2.880 \times 10^3$ J T$^{-1}$
(2) $2.880 \times 10^2$ J T$^{-1}$
(3) $2.880$ J T$^{-1}$
(4) $28.80$ J T$^{-1}$
Show Answer
Answer: (3) $2.880$ J T$^{-1}$
Solution
The neutral point is on the equatorial line. Using the equatorial field formula: $\frac{\mu_0}{4\pi} \frac{m}{r^2} \times \frac{7}{r} = 0.4 \times 10^{-4}$ $\Rightarrow 2 \times 10^{-7} \times \frac{m \times 7}{(7^2 + 18^2)^{3/2}} \times 10^4 = 0.4 \times 10^{-4}$ $M = m \times 14,\text{cm} = 4 \times 10^{-4} \times 7203.82 = 2.88$ J/T
BETA
