Q. 64.7( 23 Votes )

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?

Answer :


Current through the wire, I = 10A

Strength of magnetic field inside the solenoid = 0.27T

Angle between direction of magnetic field and current, θ = 90°

Length of the wire = 3 cm

Force on a current carrying conductor of length L carrying current I, due to a uniform magnetic field of strength B is given by,

F = IL × B …(1)

Now by plugging the values in the equation (1), we get,

F = 10A × 0.03m × 0.27T × sin90°

F = 8.1 × 10-2 N

Hence, the magnitude of force per unit length of the wire is 8.1 × 10-2 N.

Note: The direction of magnetic field inside the solenoid is parallel to its axis.

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