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

Question

Two long, parallel conductors, separated by 13.0 cm, carry currents in the same direction. The first wire carries a current I1 = 2.00 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? magnitude T direction (b) What is the force per unit length exerted by I1 on I2? magnitude N direction (c) What is the magnetic field created by I2 at the location of I1? magnitude T direction (d) What is the force per length exerted by I2 on I1? magnitude N direction

Explanation / Answer

? formula for a piece of wire: vector of magnetic induction B = (µ0*J)/(4*pi) ? [dL×r]/|r|^2, where J is current, dL is vector of elementary piece of wire, µ0=12.56637e-7, vector r is position of test point, integration runs along contour L; ? for our infinite wire the formula reduces to: B=(µ0/(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; (a)? J1=4; r=12cm=0.12m; B= 0.25*(12.56637e-7/pi) *(2*4/0.12) =6.667e-6 Tesla; (c)? J2=8; r=12cm=0.12m; B= 0.25*(12.56637e-7/pi) *(2*8/0.12) =13.333e-6 Tesla; ? a force per 1 meter of wire in magnetic field B as calculated above is dF/dL = B*J; (b)? attraction! dF/dL = 6.667e-6 *8 =53.333e-6 N; (d)? attraction! dF/dL = 13.333e-6 *4 =53.333e-6 N;

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