A 3.0 ; cm wire carrying a current of 10 ; A | vedclass

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(C) Length of the wire,$l = 3 \; cm = 0.03 \; m$.Current flowing in the wire,$I = 10 \; A$.Magnetic field,$B = 0.27 \; T$.Angle between the current an

Vedclass · Accepted Answer

$8.1 	imes 10^{-2} \; N$

Vedclass · Answer

(C) Length of the wire,$l = 3 \; cm = 0.03 \; m$.Current flowing in the wire,$I = 10 \; A$.Magnetic field,$B = 0.27 \; T$.Angle between the current and magnetic field,$	heta = 90^{\circ}$ (Because the magnetic field produced by a solenoid is along its axis and the current-carrying wire is kept perpendicular to the axis).Magnetic force exerted on the wire is given by the formula:$F = B I l \sin 	heta$Substituting the values:$F = 0.27 	imes 10 	imes 0.03 	imes \sin 90^{\circ}$$F = 0.27 	imes 10 	imes 0.03 	imes 1$$F = 8.1 	imes 10^{-2} \; N$.Hence,the magnetic force on the wire is $8.1 	imes 10^{-2} \; N$. The direction of the force can be obtained from Fleming's left-hand rule.

$A$ $3.0 \; cm$ wire carrying a current of $10 \; A$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27 \; T$. What is the magnetic force on the wire?

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