Question

Three long, parallel conductors each carry a current of I = 2.00 A. Figure P22.37 is an end view of the conductors, with each current coming out of the page. Taking a = 1.00 cm, determine the magnitude and direction of the magnetic field at (A) point A, (b) point B, and (c) point C.

Answer 1

Let the point on current is PQ&R the direction of may field at 'A' will be as shown m adjacent figure, B_p = B_a = B = mu_0 I2 pi (a squareroot 2) B_Q = mu_0 I2pi (3a) B_net at A = 2 B cos 45 + B_R = 2 times mu_0 I/2 pi a 5_2 times 1/squareroot 2 + mu_0 I/6 pi a = mu_0 I 2 pi a (1 + 1/3) 2/3 mu_0 I/3 pi a = 2/3 times 4 pi times 10^-7/pi times 2.00/1.00 times 10^-2 = 5.33 times 10^-5 (T) B_p = B_R and opposite in direction so cancelling each other so B_net at B = B_R = mu_0 I/2 pi(2a) = 4 pi times 10^-7 times 2.00/2 pi times 2 times (1.00 times 10^-2) = 2.00 times 10^-5 Heat B_p = B_a = B_R/squareroot 2 = mu_0 i/2 pi a squareroot 2 B_net = 2 B_p cos 45 - B_R = Bp (squareroot 2 - 1/squareroot 2) = 1/squareroot 2

Let the point on current is PQ&R the direction of may field at 'A' will be as shown m adjacent figure, B_p = B_a = B = mu_0 I2 pi (a squareroot 2) B_Q = mu_0 I2pi (3a) B_net at A = 2 B cos 45 + B_R = 2 times mu_0 I/2 pi a 5_2 times 1/squareroot 2 + mu_0 I/6 pi a = mu_0 I 2 pi a (1 + 1/3) 2/3 mu_0 I/3 pi a = 2/3 times 4 pi times 10^-7/pi times 2.00/1.00 times 10^-2 = 5.33 times 10^-5 (T) B_p = B_R and opposite in direction so cancelling each other so B_net at B = B_R = mu_0 I/2 pi(2a) = 4 pi times 10^-7 times 2.00/2 pi times 2 times (1.00 times 10^-2) = 2.00 times 10^-5 Heat B_p = B_a = B_R/squareroot 2 = mu_0 i/2 pi a squareroot 2 B_net = 2 B_p cos 45 - B_R = Bp (squareroot 2 - 1/squareroot 2) = 1/squareroot 2