Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.6
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, 1 = 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, 0 = 90°
Magnetic force exerted on the wire is given as:
F = BI/sin 0 = 0.27 x 10 x 0.03 sin90° = 8.1 x 10⁻2 N
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.