Question -
Answer -
Magnetic field strength, B =1.5 T
Radius of the cylindricalregion, r = 10 cm = 0.1 m
Current in the wire passingthrough the cylindrical region, I = 7 A
(a) If thewire intersects the axis, then the length of the wire is the diameter of thecylindrical region.
Thus, l = 2r =0.2 m
Angle between magnetic field andcurrent, θ = 90°
Magnetic force acting on the wireis given by the relation,
F = BIl sin θ
= 1.5 × 7 × 0.2 × sin 90°
= 2.1 N
Hence, a force of 2.1 N acts onthe wire in a vertically downward direction.
(b) Newlength of the wire after turning it to the Northeast-Northwest direction can begiven as: :
Angle between magnetic field andcurrent, θ = 45°
Force on the wire,
F = BIl1 sin θ
Hence, a force of 2.1 N actsvertically downward on the wire. This is independent of angleθbecause l sinθ isfixed.
(c) The wireis lowered from the axis by distance, d = 6.0 cm
Supposewire is passing perpendicularly to the axis of cylindrical magnetic field thenlowering 6 cm means displacing the wire 6 cm from its initial position towardsto end of cross sectional area.
Thus the length of wire in magneticfield will be 16 cm as AB= L =2x =16cm
Now the force,
F = iLB sin90° as the wire will beperpendicular to the magnetic field.
F= 7 × 0.16 × 1.5 =1.68 N
The direction will be given by righthand curl rule or screw rule i.e. vertically downwards.