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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Magnetic Properties of Matter

JEE Main · Physics · Magnetic Properties of Matter

A bar magnet is 10 ~cm long is kept with its north (N) -pole...

A bar magnet is 10 cm10 \mathrm{~cm} long is kept with its north (N)(N)-pole pointing north. A neutral point is formed at a distance of 15 cm15 \mathrm{~cm} from each pole. Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet isA bar magnet is 10 ~cm long is kept with its north (N) -pole pointing north. A neutral point is formed at a distance of

Options

  1. A9 A-m
  2. B6.75 A-m
  3. C27 A-m
  4. D1.35 A-m

Correct answer

B. 6.75 A-m

Step-by-step solution

 Length of magnet =10 cm=10×10−2 m,r=15×10−2 m\begin{aligned} & \text { Length of magnet }=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}, \\ & \qquad r=15 \times 10^{-2} \mathrm{~m}\end{aligned} A bar magnet is 10 ~cm long is kept with its north (N) -pole pointing north. A neutral point is formed at a distance of
OP=225−25=200 cmO P=\sqrt{225-25}=\sqrt{200} \mathrm{~cm} Since, at the neutral point, magnetic field due to the magnet is equal to BHBH=μ04π⋅M(OP2+AO2)3/20.4×10−4=10−7×M(200×10−4+25×10−4)3/20.4×10−410−7×(225×10−4)3/2=M0.4×103×10−6(225)3/2=MM=1.35 A−mB_H\begin{gathered} B_H=\frac{\mu_0}{4 \pi} \cdot \frac{M}{\left(O P^2+A O^2\right)^{3 / 2}} \\ 0.4 \times 10^{-4}=10^{-7} \times \frac{M}{\left(200 \times 10^{-4}+25 \times 10^{-4}\right)^{3 / 2}} \\ \frac{0.4 \times 10^{-4}}{10^{-7}} \times\left(225 \times 10^{-4}\right)^{3 / 2}=M \\ 0.4 \times 10^3 \times 10^{-6}(225)^{3 / 2}=M \\ M=1.35 \mathrm{~A}-\mathrm{m} \end{gathered}
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