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>> $A bar magnet has a magnetic moment of 20$

Question

A bar magnet has a magnetic moment of $200$ $A m^{2}$ . The magnet is suspended in a magnetic field of $0.30 N A^{- 1} m^{- 1}$ . The torque required to rotate the magnet from its equilibrium position through an angle of $30^{o}$ will be:

A

$30 N m$

B

$303 N m$

C

$60 N m$

D

$603 N m$

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Updated on : 2022-09-05

Solution

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Correct option is A)

Given $M = 200 A m^{2}; B = 0.30 N A^{- 1} M^{- 1}$
and $θ = 30^{o}$
We know that the torque
$τ = M \times B$
$\Rightarrow ∣ τ ∣ = M B sin θ = 200 \times 0.3 \times \frac{1}{2}$
$= 100 \times 0.3 = 30 N m$

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