Question

II Each wire segment in Fig. 21-64 has a length of 20.0 cm and carries a current of 3.00 A in the directions indi- cated. Find the magnitude and direction of the magnetic force on each segment if the magnetic field has a magni- tude of 0.400 T 45° Fig. 21-64

Answer 1

The force on the current carrying conductor is given by,

$\bold{F} = \bold{idl} \times \bold{B}$

The magnitude is given by,

$F = iLB sin \theta$

$\theta$ is the angle between the current and the magnetic field.

(a) Substituting values we get,

F = (3) (20 × 10-2) (0.4)sin0°-ON

Since magnitude is zero there is no direction.

(b) Substituting values we get,

F = (3)(20 × 10-2)(0.4) sin (45°) = 0.17 N

The direction will be outwards from the screen.

(c) Substituting values we get,

F= (3)(20× 10-2)(0.4)·sin (90°)= 0.24N

The direction will be outwards from the screen.

(d) Substituting values we get,

F = (3) (20 × 10-2)(0.4)sin( 180°) = 0N

Since magnitude is zero there is no direction.