Question

You make a seesaw by placing a 50-g magnet (whose poles' faces are 2-cm-by-2-cm squares) at one end of a 50-cm-longruler and a small 50-g metal object at the other end and balancing the whole setup by placing a pencil directly under the middle of the ruler. You know that the magnet is horizontally magnetized, but you don't know which face is north and which face is south. When you place a horizontal wire close to the pole of the magnet and send a 14-A current through the wire in the direction shown in (Figure 1), the end of the seesaw with the magnet goes up and the other end goes down. You can get the seesaw to go back to equilibrium (while keeping the current in the wire) by placing another 10-g metal object 7.0 cm from the axis of rotation, also shown in the figure.

Assume that the blocks are much smaller than 50 cm.

Magnet

A. Determine the magnitude of the B⃗  field at the location of the wire.

Answer 1

Torque produced by 50g object (lambda) = tau_A = mg times n = 0.05 times 9.8 times 0.25 tau_A = 0.1225 Nm Gaslies this torque was maintained by torque of magnet But when wire is Brought close to magnet Net force on magnet reduces F_B_Act = mg - Bil When l = 2cM Now torque is maintained by placing another object �c� 7cm from axis of solution so net torque of magnet + object (10g object) tau_Act = (mg - Bil) times 0.25 + 0.01 times 9 times 0.07 This T_Act must be equal to tau_A 0.1225 = (0.05 times 9.8 times 2.5 - 8 times 14 times 0.02 times 0.25 + 0.01 times 9.8 times 0.07 0.1225 = (0.1225 - 0.07B + 0.00686 0.1225 = 0.12936 - 0.07B 0.07B = 0.12936 - 0.1225 B = 0.00686/0.07 B = 0.098 Tesla