Question

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 1

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, \(θ = 90° \)
Magnetic force exerted on the wire is given as:
 \(F= BISinθ\)
\(=0.27 × 10 × 0.03 \,sin90°\)
\(= 8.1 × 10^{–2} N\)
Hence, the magnetic force on the wire is \(8.1 × 10^{–2} N\). The direction of the force can be obtained from Fleming’s left hand rule.