A 3.0 cm wire carryinga current of 10 A is placed inside a solenoid perpendicular to itsaxis. 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 thewire, I = 10 A
Magnetic field, B= 0.27 T
Angle between thecurrent and magnetic field, θ = 90°
Magnetic force exertedon the wire is given as:
F= BIlsinθ
=0.27 × 10 × 0.03 sin90°
=8.1 × 10–2N
Hence, the magneticforce on the wire is 8.1 × 10–2 N. Thedirection of the force can be obtained from Fleming’s left handrule.