a) what is the force on section ab? b) what is the force on section cd? c) what

Question

a) what is the force on section ab?

b) what is the force on section cd?

c) what is teh force per unit length on section bc?

d) Assume that due to the forces in parts a and c the loop spins about its centeral axis, half way between points and and d, what is the torque on the current loop causing it to spin?

The current in the straight wire is I1, is 10 Amperes. The current in the loop, I2, is 2 Amperes. The distance from I1 to section dc is 50 cm, the distance to ab is 100 cm, and section dc and ab are 1 meter long. a) what is the force on section ab? b) what is the force on section cd? c) what is teh force per unit length on section bc? d) Assume that due to the forces in parts a and c the loop spins about its centeral axis, half way between points and and d, what is the torque on the current loop causing it to spin?

Explanation / Answer

a) To find force on section ab, we need to find magnetic field due to wire having current I1. Formula to be used is-

Bab = µo I1 / 2 dab = 4 * 10-7 * 10 / 2 * 1 = 2 * 10-6 T.

Fab = I1.Bab. lab = 2 * 2 * 10-6 * 1 = 4 * 10-6 N.

b) Fcd = I1.Bcd. lcd = 2 * 4 * 10-6 * 1 = 8 * 10-6 N.

c) Since bc is perpendicular to wire carrying I1 current, there will be zero magnetic field experienced by bc and thus no force.

Force on bc/unit length = Zero

d) When the loop spins half way, then distance between I1 wire and loop would be 0.5 m.

Torque = I.A.B = 2 * [1 * 0.5] * [4 * 10-7 *10 / 2 * 0.5] = 4 * 10-6 Nm.

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