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

Magnetic field strength `=B = 1.5T`
radius of the cylindrical region, `r = 10 cm = 0.1m`
Current in the wire passing through the cylindrical region, `I = 7A`
(a) If the wire intersects the axis, then the length of the wires is the diameter of the cylindrical region.
Thus, `l = 2r = 0.2m`
Angle between magnetic field and current, `theta = 90^(@)`
Magnetic force acting on the wire is given by the relation,
`F = BI sin theta`
`= 1.5 xx 7 xx 0.2 xx sin 90^(@)`
`=2.1N`
Hence, a force of `2.1N` acts on the wire in a vertically downward direcion.
(b) New length of the wire after turning it to the Northeast-Northwest direction can be given as:
`l_(1) = (l)/(sin theta)`
Angle between magnetic field and current, `theta = 45^(@)`
Force on the wire,
`F = BIl_(2) sin theta`
`=BIl`
`=1.5 xx 7 xx 0.2`
`=2.1N`
Hence, a force of `2.1N` acts vertically downward on the wire. This is independent of angle `theta` because `l sin theta` is fixed.
(c) The wire is lowered from the axis by distance, `d = 6.0cm` Let `l_(2)` be the new length of the wire.
`:. ((l_(2))/(2))^(2) = 4(d+r)`
`=4 (10+6) =4(16)`
`:. l_(2) = 8 xx 2 = 16cm = 0.16m`
Magnetic force exerted on the wire,
`F_(2) = BIl_(2)`
`=1.5 xx 7 xx 0.16`
`= 1.68 N`
Hence, a force of `1.68N` acts in a vertically downward direction on the wire.