Question

A uniform 2.5 T magnetic field points to the right. A 3.0-m-long wire, carrying 15 A, is placed at an angle of 30 degree circ to the field, as shown in the figure.

What is the magnitude of the force on the wire?

Answer 1

All your information is contained in the magnetic force equation which is a vector cross product between the magnetic field, B , and the current element , iL , as;

F = iL x B

This has magnitude;

F = iLBSin(a) , where , a , is the angle between iL & B

= (15)(3)(2.5)Sin(30) = 56.25 N

The direction is perpendicular to the plane formed by vectors iL & B and is found by curling fingers from iL toward B and your thumb points in direction of the force. (right hand rule)

Answer 2

F = Bil sin30 = 2.5*3*15*0.5 =56.25 N

Answer 3

Magnitude of force on wire = B*I*L*sin(θ) = 2.5*15*3*sin(30⁰) = 56.25N

Answer 4

F = iL x B.................F = iLBSin(a) .........56.25 N