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Question: The magnetic moment of a short bar magnet placed with its magnetic axis at 30° to an external field ...

The magnetic moment of a short bar magnet placed with its magnetic axis at 30° to an external field of 900 G and experiences a torque of 0.02 N m is

A

0.35 Am2

B

0.44 Am2

C

2.45 Am2

D

1.5 Am2

Answer

0.44 Am2

Explanation

Solution

: Here, B = 900 Gauss = 900×10−4 T900 \times 10 ^ { - 4 } \mathrm {~T}

=9×10−2 T= 9 \times 10 ^ { - 2 } \mathrm {~T}

θ=30∘\theta = 30 ^ { \circ }

∴\therefore

⇒\Rightarrow 0.02=m×9×10−2×sin⁡30∘0.02 = \mathrm { m } \times 9 \times 10 ^ { - 2 } \times \sin 30 ^ { \circ }

m=0.02×29×10−2=0.44Am2\mathrm { m } = \frac { 0.02 \times 2 } { 9 \times 10 ^ { - 2 } } = 0.44 \mathrm { Am } ^ { 2 }