A coil 2.0 cm in diameter has 300 turns. If the coil carries a current of 10 mA and lies in a magnetic field 5 × $10^{- 2}$ T, the maximum torque experienced by the coil is :

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$4.7 \times 10^{- 2} N - m$

$4.7 \times 10^{- 4} N - m$

$4.7 \times 10^{- 5} N - m$

$4.7 \times 10^{- 8} N - m$

answer is C.

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Detailed Solution

$\begin{aligned} τ_{max} = NiAB \\ τ_{max} = 300.10 \times 10^{- 3} \times \frac{π 4 \times 10^{- 4}}{4} \times 5 \times 10^{- 2} = 4.7 \times 10^{- 5} \end{aligned}$

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