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Question: A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetisation...

A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetisation of 5.30 × 103Amp/m3. What its magnetic dipole moment

A

1×10−2J/T1 \times 10 ^ { - 2 } J / T

B

C

3.08×10−2 J/T3.08 \times 10 ^ { - 2 } \mathrm {~J} / \mathrm { T }

D

Answer

Explanation

Solution

Relation for dipole moment is, M=I×VM = I \times V , Volume of the cylinder V=πr2lV = \pi r ^ { 2 } l Where r is the radius and l is the length of the cylinder, then dipole moment,

M=I×πr2l=(5.30×103)×227×(0.5×10−2)2(5×10−2)M = I \times \pi r ^ { 2 } l = \left( 5.30 \times 10 ^ { 3 } \right) \times \frac { 22 } { 7 } \times \left( 0.5 \times 10 ^ { - 2 } \right) ^ { 2 } \left( 5 \times 10 ^ { - 2 } \right)