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 `1=3cm=0.03m,, ` Current flowing in the wire `1-10A`
Magnetic field `B=0.27T`
Angle between the current and magnetic field,`theta=90^(@)` (Because magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis)
`F=B i l sin theta, =0.27xx10xx0.03sin 90^(@), =8.1xx10^(-2)N`
Hence, the magnetic force on the wire is `8.1xx10^(-2)N`. The direction of the force can be obtained from Fleming’s left hand rule