A 15 A current-carrying wire has a length of 27 cm in a magnetic field of 0.79 T.

Question

A 15 A current-carrying wire has a length of 27 cm in a magnetic field of 0.79 T.

Find the force on the wire when it makes the following angles with the magnetic field lines of

a. 90°

b. 45°

c. 180°

d. 0°

Answer 1

The equation for this problem is F = I x l X B where I X B is the cross product of the wire length and the magnetic field strength (assuming the wire is straight). The cross product is the length of the wire times the magnetic field strength time the sine of the angle between the wire and field strength vectors.

So the force F = I x l (length) x B sin θ

a) F = 15 A x 0.27 m x 0.79T x sin 90° = 3.2 newtons

b) to d) are done similarly by changing the angle.

Answer 2

This problem has to do with the Current in a magnetic field that causes a force on the wire.

F=I*length*B if the wire is perpendicular to the magnetic field.

F= I *length*B * sin (angle)

a For 90 degrees F=I* length* B * sin(90) = 0.27 m * 15 A * 0.79 T *1 = 3.200N

b For 45 degrees F=I * length*B*sin (45) = 0.27 m *15 A * 0.79 T * 0.707 = 2.262 N

In c) andD) sin(angle) =0 so the Force F = 0 N.

Besure to change the units on the length = 27 cm.