COMEDK · Physics · 20. Magnetic Properties of Matter
A bar magnet of length 6 cm is placed in the magnetic meridian with N pole pointing towards the geographical north. Two neutral points, separated by a distance of 8 cm are obtained on the equatorial axis of the magnet. If \(B_H = 1.2 \times 10^{-5}\text{ T}\), then the pole strength of the magnet is:
- A \(0.75\text{ A}\cdot\text{m}\)
- B \(0.25\text{ A}\cdot\text{m}\)
- C \(50\text{ A}\cdot\text{m}\)
- D \(1.50\text{ A}\cdot\text{m}\)
Answer & Solution
Correct Answer
(B) \(0.25\text{ A}\cdot\text{m}\)
Step-by-step Solution
Detailed explanation
Given: \(2l = 6\,\text{cm} \Rightarrow l = 0.03\,\text{m}\), \(d = \dfrac{8}{2} = 4\,\text{cm} = 0.04\,\text{m}\), \(B_H = 1.2 \times 10^{-5}\,\text{T}\) At neutral point on equatorial axis: \(B = B_H\) \(\dfrac{\mu_0}{4\pi} \cdot \dfrac{m \times 2l}{(d^2 + l^2)^{3/2}} = B_H\)…
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