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The torque required to keep a magnet of length 20 cm at 30º to a uniform field is 2×10^–5N-m. The magnetic force on each pole is

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2 × 10^–4 N

4 × 10^–4 N

2 × 10^–3 N

2 × 10^–6 N

answer is A.

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Detailed Solution

$τ = M B s i n 30^{0}$

$\Rightarrow 2 \times 10^{- 5} N = m \times 0.20 \times B \times \frac{1}{2}$

$\Rightarrow m B = F = 2 \times 10^{- 4} N$

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