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

The magnitude of magnetic force acting on a current carrying wire of length l and carrying current I in an external magnetic field B and making an angle `theta` with the direction of magnetic field is given by:
` F = BI l sin theta`
Here, the external magnetic field is that of the solenoid. Given that the wire is placed inside the solenoid perpendicular to its axis. Since the magnetic field due to a solenoid is along the axis of solenoid, this means that the angle `theta = 90^(@)`
Also, `l = 3.0 cm = 3 xx 10^(-2) m`
` I = 10 A`
` B = 0.27 T`
` rArr" " F = B I l sin theta`
` = 0.27 xx 10 xx (3 xx 10^(-2)) sin 90^(@)`
` = 8.1 xx 10^(-2) N`
To find the direction of magnetic force on the wire, we will apply right-hand palm rule. According to this rule, the direction of magnetic force will be perpendicular to the plane of paper and inwards.