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?
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, Ω¸ = 90°
Magnetic force exerted on the wire is given as:
F= BIlsinΩ¸
= 0.27 x 10 x 0.03 sin90°
= 8.1 x 10-2N
Hence, the magnetic force on the wire is 8.1 x 10-2 N. The direction of the force can be obtained from Fleming's left hand rule.
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