Question

A wire having a mass per unit length of 0.480 g/cm carries a 2.50-A current horizontally to the south.

(a) What is the direction of the minimum magnetic field needed to lift this wire vertically upward?

EASTWARD (correct)

(b) What is the magnitude of the minimum magnetic field needed to lift this wire vertically upward?
__________ T

Answer 1

a) The directions of the magnetic field, current and the force experienced by the conductor are mutually perpendicular and given by right hand rule. The thumb shows force direction, the pointing fingers shows current and the middle finger points to magnetic field when all held at right angles to each other. Thus, keeping thumb upwards, pointing finger towards South, we see middle finger towards east. Thus, eastwards is the answer.

b) Now, F = I*L*B*sin©, thus, F/L = I*B*sin©.... Eq(1)

where © is the angle between current and magnetic field direction. In this case, © = 90°, sin© = 1.

Now, to lift the wire we need force equal to its weight.

Mass per unit length = M/L = 0.48 g/cm = 0.048 kg/m.

Weight per unit length = W/L = 0.048*9.8 = 0.4704 N/m.

Thus, we require F/L = 0.4704 N/m to lift the wire.

Thus, from EQ(1),

0.4704 = I*B*(1), I = 2.5A, thus, B = 0.4704/2.5 = 0.188 T

B= 0.188T