Question

A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of 0.22 m and a mass of 0.009 kg. A uniform magnetic field of magnitude 0.075 T is directed from the ceiling to the floor. When a current of I = 31 A exists in the wire, the wire swings upward and, at equilibrium, makes an angle φ with respect to the vertical, as the drawing shows. Find (a) the angle and (b) the tension in each of the two strings.

φ =_______ degrees

T= _______ N

Answer 1

Looking down the wire we have weight (m*g) acting down, The FB (force from the magnetic field) to the rt = I*L*B and the tension T in the strings acting along the strings. (Due to symmetry the tension in each string is the same)

Now sum forces in the vertical we have 2*T*cos(φ) = m*g
And in the horizontal I*L*B = 2*T*sin(φ)

So dividing the 2nd eqn by the first to eliminate T

we get sin(φ)/cos(φ) = tan(φ) = I*L*B/m*g = 31*0.22*0.075/(0.009*9.8) = 5.7993

so φ = arctan(5.7993) = 80.216o

b) therefore the tension in each string is T = m*g/(2*cos(φ)) = 0.009*9.8/(2*cos(81.216)) = 0.2887N