Question

A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A 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 axis
(b) The wire is turned from N-S to northeast-northwest direction,
(c) The wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Answer 1

F = BIsinθ
B = 1.5 T
I = 7A
θ = 90°
Radius of the cylindrical region of magnetic field, r = 10 cm = 0.1 m
The length AB of the wire inside the magnetic field,
1 = 2 x radius
= 2 x 0.1m = 0.2 m
F=1.5 x 7 x 0.2 x sin90°
= 1.5 x 7 x 0.2
= 2-1 N (vertically downwards)

This figure shows the change in orientation of the wire. It follows that now the angle between the direction of magnetic field and the direction of flow current will be 45°. The length AB of the wire inside the magnetic field,

The length AB of the wire inside the magnetic field can be found from the chord property of the circle. It follows that AC x BC = PC × QC

= 1.5 x 7 x 0.16 x sin90°
= 1.5 x 7 x 0.16
= 1.68 N (vertically downwards).