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?
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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