Question

A single-turn square loop carries a current of 16 A . The loop is 15 cm on a side and has a mass of 3.8×10⁻² kg . Initially the loop lies flat on a horizontal tabletop. When a horizontal magnetic field is turned on, it is found that only one side of the loop experiences an upward force.

Find the minimum magnetic field, Bmin, necessary to start tipping the loop up from the table. Express your answer using two significant figures.

Answer 1

magnetic force on a current carrying conductor

Fm = i l X B , here l and B are perpendicular

i = 16 A, l = 0.15 m

Fm = 16*0.15*B = 2.4B N

this force will cause a torque on the loop about its other edge

mass of the loop = 3.8e-2 kg

weight of the loop 3.8e-2 *g acts downward at the center of the loop and causes CW torque , whereas the magnetic force Fm causes CCW torque. weight of the loop acts a distance (0.15/2) m,and the magnetic force Fm at distance of 0.15 m about the other edge. for the loop to tip up CCW torque must be more than the CW torque

2.4B * 0.15 > 0.15/2 * 3.8e-2 *g

B > 77.85 mT

the minimum field required to tip up the loop = 77.5 mT