A closely wound solenoid of $800$ turns and area of cross section $2.5 \times 10^{- 4} m^{2}$ carries a current of $3 A$ . Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

0.9 J T^{- 1}

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0.6 J T^{- 1}

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0.7 J T^{- 1}

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0.8 J T^{- 1}

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Solution

The correct option is B $0.6 J T^{- 1}$
Given, number of turns of the solenoid, N= 800
Loop area, A= $2.5 \times 10^{- 4} m^{2}$
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis.
The magnetic moment associated with the solenoid is, m= NIA
m= $= 800 \times 3 \times 2.5 \times 10^{- 4} J / T$
$= 0.6 J T^{- 1}$

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