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Question: A bar magnet of magnetic moment 3.0 A-m<sup>2</sup> is placed in a uniform magnetic induction field ...

A bar magnet of magnetic moment 3.0 A-m2 is placed in a uniform magnetic induction field of 2 × 10–5 T. If each pole of the magnet experiences a force of 6 × 10–4 N, the length of the magnet is

A

0.5 m

B

0.3 m

C

0.2 m

D

0.1 m

Answer

0.1 m

Explanation

Solution

F=mB⇒F=ML×BF = m B \Rightarrow F = \frac { M } { L } \times B

⇒6×10−4=3L×2×10−5⇒L=0.1 m\Rightarrow 6 \times 10 ^ { - 4 } = \frac { 3 } { L } \times 2 \times 10 ^ { - 5 } \Rightarrow L = 0.1 \mathrm {~m}