What will be the force on a segment of wire 11 cm long, carrying a current of 13 amps, in the presence of a .0009 Tesla magnetic field directed perpendicular to the segment? If the current is directed toward the North and the magnetic field vertically upward, what will be the direction of this force?
The magnitude of the force on a charge q moving with velocity v in a direction perpendicular to a magnetic field B is F = (q * v) * B. From this it can be concluded that the force on a current I in a wire of length L is F = ( I * L ) * B. (You might try showing that the sum of all the q * v contributions from current I in length L of a wire is I * L).
In the present case the current is moving perpendicular to the field, so that the force is
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, which is directed toward the North, with the palm facing vertically upward in the direction of the magnetic field. The thumb will therefore be pointing East, and the force will be toward the East.
In general the force on a charge q moving at velocity v perpendicular to a magnetic field B is F = q v B.
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 experienced by these charges is