a current-carrying wire is oriented at right angles to a uniform magnetic field. if the lenght of the wire is 0.50m and it experiences a force at 0.90n while carrying a current of 4.0a. what is the strength of the magnetic field?

Answer 1

The answer is .45 T
F = B(I)L
B = F/(I)L
B = .90/4(.5)
B = .45 T
(T is Teslas)