Two long, parallel conductors, separated by 12.0 cm, carry currents in the same

Question

Two long, parallel conductors, separated by 12.0 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?

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

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

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

magnitude T direction ---Select--- 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

Formula for a piece of wire: vector of magnetic induction

B = (0*J)/(4*pi) [dL×r]/|r|^2, where J is current, dL is vector of elementary piece of wire, 0=12.56637e-7, vector r is position of test point, integration runs along contour L;

for our infinite wire the formula reduces to:
B = (0/(4*pi)) *(2*J/r), where r is distance from the wire, direction of B is defined according to skrew rule for direction of J;

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(a) J1=6; r=12cm=0.12m;
B= 0.25*(12.56637e-7/pi) *(2*6/0.12) =9.999e-6 T;

This B field depends on your distance from I1, but because the wires are parallel, the B field from I1 is constant along I2 We can use the right hand rule to determine that B1 is perpendicular to both I1 and r.

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(b) attraction dF/dL = 9.999e-6 *8 =7.99999e-5 N;

where there is no sin term in the cross product, because B1 is perpendicular to I2. By drawing the situation and doing some right hand rules, you can convince yourself that this force is attractive.

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(c) Because the situation in (c) is identical to (a)

J2=8; r=12cm=0.12m;
B= 0.25*(12.56637e-7/pi) *(2*8/0.12) =13.333e-6 T;
  

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(d) attraction dF/dL = 13.333e-6 *6 =7.999999e-5 N;

as we would expect from Newton’s third law (for every action there is an equal and opposite reaction).

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