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 = 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
What is the equation giving the magnetic field of a long, straight, current-carrying wire? T direction

Explanation / Answer

magnetic field by a current wire is given by

B = u0*I/(2*pi*r)

a)

magnetic field at the location of I2

B = u0*I1(2*pi*0.13) = (4*pi*10^(-7)*6)/(2*pi*0.13) = 9.23*10^(-6) T

and direction is out of the page from right hand thumb rule.

b)

magnetic force per unit length exerted by I1 on I2 is given by

F/L = u0*I1*I2/(2*pi*r) = (4*pi*10^(-7)*6*8)/(2*pi*0.13) = 7.38*10^(-5) N/m

and direction is out of the page.

c)

magnetic field at the location of I1 due to I2

B = u0*I2(2*pi*0.13) = (4*pi*10^(-7)*8)/(2*pi*0.13) =12.3*10^(-6) T

and direction is into the page from right hand thumb rule.

d)

magnetic force per unit length exerted by I2 on I1 is given by

F/L = u0*I1*I2/(2*pi*r) = (4*pi*10^(-7)*6*8)/(2*pi*0.13) = 7.38*10^(-5) N/m

and direction is into the page.

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