A uniform magnetic field of `1*5T` is in cylindrical region of radius `10*0cm` with its direction parallel to the axis along east to west. A wire carrying current of `7*0A` in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if (a) the wire intersects the axes, (b) the wire is turned from N-S to north east-south west direction, (c) the wire in the N-S direction is lowered from the axis by a distance `6*0cm`?

Explanation

Force acting on a current carrying wire of length l and carrying a current l when placed in an external magnetic field B is given by:
`F = Bil sin theta`,
where `theta` is the angle between wire and the magnetic field

Here, `B = 1.5 T`
Length of the wire in the cylindrical region i.e. equal to the diameter of cylindrical region ` = 2r`
` = 2 xx 10 cm = 20 = 0.2 m`
` I = 7A`
` theta = 90^(@)`
` :. " " F = 1.5 xx 7 xx 20/100`
` rArr " " F = 2.1 N `
The direction of magnetic force can be found by using Fleming.s left-hand rule. Since magnetic field is from east to west and current in the wire is flowing from north to south, the force will be acting vertically downwards.
(b)

Now `theta = 45^(@)`, and length of the wire in the cylindrical region of the magnetic field is `l_(1)` (say) and is given by:
` l = l_(1) sin 45^(@)`
The force acting on the wire in this case will be
` F_(1) = BIl_(1) sin 45^(@)`
` = BIl`
` = 1.5 xx 7.0 xx 0.2`
` = 2.1` N ( vertically downwards)
(c ) When the wire is lowered through a distance `6.0` cm, and reaches at CD then length of the wire in the magnetic field,
` l_(2) = 2x`
and x is given by
` x^(2) = 10^(2) - 6^(2)`
` rArr" " x = 8 cm`
` :. " " l_(2) = 8 xx 2 = 16 cm`
` = 16/100 m`

`:. ` Force on the wire
` F_(2) = BI l_(2)`
` = (1.5 xx 7 xx 16)/100`
` = 1.68 N` ( vertically downwards)