A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm,

Solution:

Magnetic field strength, B = 1.5 T

Radius of the cylindrical region, r = 10 cm = 0.1 m

Current in the wire passing through the cylindrical region, I = 7 A

(a) If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region.

Thus, l = 2r = 0.2 m

Angle between magnetic field and current, θ = 90°

Magnetic force acting on the wire is 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 on the wire in a vertically downward direction.

(b) New length of the wire after turning it to the Northeast-Northwest direction can be given as: :

$l_{1}=\frac{l}{\sin \theta}$

Angle between magnetic field and current, θ = 45°

Force on the wire,

F = BIl1 sin θ

$=B I l$

$=1.5 \times 7 \times 0.2$

$=2.1 \mathrm{~N}$

Hence, a force of 2.1 N acts vertically downward on the wire. This is independent of angleθbecause l sinθ is fixed.

(c) The wire is lowered from the axis by distance, d = 6.0 cm

Suppose wire is passing perpendicularly to the axis of cylindrical magnetic field then lowering 6 cm means displacing the wire 6 cm from its initial position towards to end of cross sectional area. 

Thus the length of wire in magnetic field will be 16 cm as AB= L =2x =16 cm

Now the force,

F = iLB sin90°  as the wire will be perpendicular to the magnetic field.

F= 7 × 0.16 × 1.5 =1.68 N

The direction will be given by right hand curl rule or screw rule i.e. vertically downwards.