Question

We want to find the magnetic field at one corner of a square; the field is produced by currents that are located at the other three corners of the square. (See drawing.) The square is in the x-y plane and the currents are flowing parallel to the z axis. We define a positive current as one flowing in the + k direction. 13 The first current, of 7.15 A, is located 3.03 cm along the x axis, that is, at the point (3.03,0.00) cm. The second current, of -2.10 A, is located at the far corner, that is, at the point (3.03,3.03) cm. The third current, of 7.15 A, is located 3.03 cm along the y axis, that is, at the point (0.00,3.03) cm. Find the magnetic field at the origin. Enter the x and y components of the field: ( 4.028E-5T i + 4.028E-5T Draw a careful diagram and pay attention to the r.h. rule and the direction of each current. i Submit Answer Incorrect. Tries 1/99 Previous Tries

Answer 1

given
I1 = 7.15 A
I2 = -2.10 A
I3 = 7.15 A
d = 3.03 cm 0.0303 m

at origin magnitude of magnetic field due to I1,
B1 = mue*I1/(2*pi*d)

= 4*pi*10^-7*7.15/(2*pi*0.0303)

= 4.72*10^-5 T

at origin magnitude of magnetic field due to I2,
B2 = mue*I2/(2*pi*sqrt(2)*d)

= 4*pi*10^-7*2.1/(2*pi*sqrt(2)*0.0303)

= 0.980*10^-5 T

at origin magnitude of magnetic field due to I3,
B3 = mue*I3/(2*pi*d)

= 4*pi*10^-7*7.15/(2*pi*0.0303)

= 4.72*10^-5 T

x-compoment of net magneti field,

Bnetx = B1x + B2x + B3x

= 0 - 0.980*10^-5*cos(45) + 4.72*10^-5

= 4.03*10^-5 T

y-compoment of net magneti field,

Bnety = B1y + B2y + B3y

= -4.72*10^-5 + 0.980*10^-5*cos(45) + 0

= -4.03*10^-5 T

so, net magnetic field at origin,

Bnet = Bnetx i + Bnety j

= 4.03*10^-5 i - 4.03*10^-5 j <<<<<<----------Answer

Note : please comment of any further clarification.