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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Length of the wire, $l=3 \,c m=0.03 \,m$

Current flowing in the wire, $I=10 \,A$

Magnetic field, $B =0.27 \,T$

Angle between the current and magnetic field, $\theta=90^{\circ}$ ( Because magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis)

Magnetic force exerted on the wire is given as

$F=B I \sin \theta$

$=0.27 \times 10 \times 0.03 \sin 90^{\circ}$

$=8.1 \times 10^{-2} \;N$

Hence, the magnetic force on the wire is $8.1 \times 10^{-2} \,N$. The direction of the force can be obtained from Fleming's left hand rule.

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