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Two long, parallel conductors, separated by 13.0 cm, carry currents in the same

ID: 1391301 • Letter: T

Question

Two long, parallel conductors, separated by 13.0 cm, carry currents in the same direction. The first wire carries a current I1 = 6.00 A, and the second carries I2 = 8.00 A. (See figure below. Assume the conductors lie in the plane of the page.)

(a) What is the magnetic field created by I1 at the location of I2?

in the +x direction

in the -x direction

in the +y direction

in the -y direction

in the +z direction

in the -z direction


(b) What is the force per unit length exerted by I1 on I2?

in the +x direction

in the -x direction

in the +y direction

in the -y direction

in the +z direction

in the -z direction


(c) What is the magnetic field created by I2 at the location of I1?

in the +x direction

in the -x direction

in the +y direction

in the -y direction

in the +z direction

in the -z direction


(d) What is the force per length exerted by I2 on I1?

magnitude ? T direction

in the +x direction

in the -x direction

in the +y direction

in the -y direction

in the +z direction

in the -z direction

Explanation / Answer

a)
magnetic field due to I1 is

B1 = u0 * I1/(2*pi*d)

B1 = u0 * 6 /(2 *pi*0.13)

B1 = 9.23 *10^-6 T in +ve Z - direction

b)

force on I2 by I1 ,

F = B*I2

F = 9.23 *10^-6 * 8

F = 7.4 *10^-5 N/m

the force on I2 is 7.4 *10^-5 N/m in +y - direction

c)

Now, field due to I2 ,

B2 = u0 * I2/(2*pi*d)

B2 = u0 * 8 /(2 *pi*0.13)

B2 = 1.231 *10^-5 T

field due to wire 2 is 1.231 *10^-5 T in -ve Z - direction

d)

Using newton's third law ,


F = 7.4 *10^-5 N/m

the force on I2 is 7.4 *10^-5 N/m in - y - direction

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