A 3.0 \mathrm{~cm} wire carrying a current of 10 \mathrm{~A} is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 \mathrm{~T}. What is the magnetic force on the wire?
A 3.0 \mathrm{~cm} wire carrying a current of 10 \mathrm{~A} is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 \mathrm{~T}. What is the magnetic force on the wire?

Ans: We are given the following:
Length of the wire, I=3 \mathrm{~cm}=0.03 \mathrm{~m}
Current flowing in the wire, 1=10 \mathrm{~A}.
Magnetic field, B=0.27 \mathrm{~T}
Angle between the current and magnetic field, \theta=90^{\circ}
Magnetic force exerted on the wire could be given as:
F=B \| \sin \theta
Substituting the given values, we get,

    \[\begin{array}{l} \Rightarrow F=0.27 \times 10 \times 0.03 \sin 90^{\circ} \\ \Rightarrow F=8.1 \times 10^{-2} \mathrm{~N} \end{array}\]

Therefore, the magnetic force on the wire is found to be 8.1 \times 10^{-2} \mathrm{~N} and the direction of the force can be obtained using Fleming’s left-hand rule.