AP EAMCET · PHYSICS · Magnetic Properties of Matter
A bar magnet is \(10 \mathrm{~cm}\) long is kept with its north \((N)\)-pole pointing north. A neutral point is formed at a distance of \(15 \mathrm{~cm}\) from each pole. Given the horizontal component of earth's field is 0.4 Gauss, the pole strength of the magnet is
- A 9 A-m
- B 6.75 A-m
- C 27 A-m
- D 1.35 A-m
Answer & Solution
Correct Answer
(B) 6.75 A-m
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Length of magnet }=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}, \\ & \qquad r=15 \times 10^{-2} \mathrm{~m}\end{aligned}\) \(O P=\sqrt{225-25}=\sqrt{200} \mathrm{~cm}\) Since, at the neutral point, magnetic field due to the magnet is equal to…
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