A bar magnet of pole strength 2 A-m is kept in a magnetic field of induction 4×10–5wbm–2 such that the axis of the magnet makes an angle 300 with the direction of the field. The couple acting on the magnet is found 80×10–7 N-m. Then the distance between the poles of the magnet is

2 ℓ = τ mBsin ⁡ θ = 80 × 10 − 7 2 × 4 × 10 − 5 × 1 2 = 20 × 10 − 2 m 2 l = 20 cm

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A bar magnet of pole strength 2 A-m is kept in a magnetic field of induction 4×10^–5wbm^–2 such that the axis of the magnet makes an angle 30⁰ with the direction of the field. The couple acting on the magnet is found 80×10^–7 N-m. Then the distance between the poles of the magnet is

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20 cm

20 m

3 cm

answer is D.

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Detailed Solution

$\begin{array}{l} 2 ℓ = \frac{τ}{mBsin θ} = \frac{80 \times 10^{- 7}}{2 \times 4 \times 10^{- 5} \times \frac{1}{2}} = 20 \times 10^{- 2} m \\ 2 l = 20 cm \end{array}$

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