A segment of wire is 15 cm long and carries a current of 15 amps. If the current is directed along the positive x axis and a magnetic of .0072 Tesla is directed at 56 degrees with the z axis, with the field vector parallel to the y-z plane, then what will be the magnitude and the direction of the force exerted on the wire by the field?
The magnitude of the force on a charge q moving with velocity v in a direction which makes angle `theta with a magnetic field B is F = (q * v) * B * sin(`theta). From this it can be concluded that the force on a current I in a wire of length L is F = ( I * L ) * B * sin(`theta). (the sum of all the q * v contributions in the wire is easily shown to be I * L).
It the current was moving perpendicular to the field, sin(`theta) would be 1 and the force would be
Since the current and field are at an angle of 56 degrees, the force will be
The direction of the force is found from the right-hand rule by 'turning' the velocity vector toward the field vector and checking the direction of the thumb. The fingers will point along the velocity vector, is in the x direction, with the palm turned so it faces the direction of the velocity vector (imagine holding the palm so it faces vertically upward, then turning it 56 degrees toward the y axis). The thumb will point 56 degrees below the y axis, in the y-z plane, and this will be the direction of the force vector.
In general the force on a charge q moving at velocity v at an angle `theta with respect to a magnetic field B is F = q v B sin(`theta).
A current I in a straight wire segment of length L is equivalent to charge q = I `dt traveling distance L in time `dt, therefore having velocity v = L / `dt, so that q v = I `dt * L / `dt = I * L. It immediately follows that the force F = q v B sin(`theta) experienced by these charges is