A closely wound solenoid of $800$ turns and area of cross section $2.5 \times 10^{-4} \;m ^{2}$ carries a current of $3.0\; A .$ Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
A closely wound solenoid of $800$ turns and area of cross section $2.5 \times 10^{-4} \;m ^{2}$ carries a current of $3.0\; A .$ Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Number of turns in the solenoid, $n=800$
Area of cross-section, $A=2.5 \times 10^{-4} \,m ^{2}$
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, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
$M=n I A$
$=800 \times 3 \times 2.5 \times 10^{-4}$
$=0.6 \,J\,T ^{-1}$
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