A `3*0cm` wire carrying a current of `10A` is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be `0*27T`. What is the magnetic force on the wire?

Explanation

Length of the wire, `l =3cm = 0.03m`
Current flowing in the wire, `I = 10A`
Magnetic field, `B = 0.27T`
Angle between the current and magnetic field `theta = 90^(@)`
Magnetic force exerted on the wire is given as:
`F = BIl sin theta`
`=0.27 xx 10 xx 0.03 sin 90^(@)`
`= 8.1 xx 10^(-2)N`
Hence, the magnetic force on the wire is `8.1 xx 10^(-2)N`. The direction of the force can be obtained from Fleming's left hand rule.