The diagram above shows segments of two long straight wires, carrying currents I
ID: 1787642 • Letter: T
Question
The diagram above shows segments of two long straight wires, carrying currents I1 = 4.0 and I2 = 3.4 respectively. The distance between the wires is 6.8 cm.
A) What is the magnitude of the force per unit length between these two wires – both magnitude and direction? Answer: ______________.___x 10 N/m
B) How far to the right of wire I1 is the total magnetic field due to both currents equal to zero? Answer: _____________.___ cm
C) Now consider the charge q = 3.1 x 10 C, located a distance of 2.9 cm to the right of wire I2, moving to the right at speed v = 46.6 m/s. What is the magnitude of the total magnetic force on this charge? Answer: ____________.___x 10 N
SECTION 1 I1 12Explanation / Answer
Given
I1 = 4.0 A and I2 = 3.4 A
separation of the wires is r = 6.8 cm
3.4
A)
magnitude of the magnetic force is F = mue0*I1*I2 /(2pi*r)
F = 4pi*10^-7*4*3.4/(2pi*0.068) N = 4*10^-5 N
the direction of the force per unit length is towards each other
B)
How far to the right of wire I1 is the total magnetic field due to both currents equal to zero?
let the point between the wires is d from I1 and r-d from I2
the magnetic field due to I1 is , B1 = mue0*I1 /(2pi*d) = 4pi*10^-7*4/(2pi*d)
and which is equal to the magnetic field due to I2 , B2 = mue0*I2 /(2pi*(r-d)) = 4pi*10^-7*3.4/(2pi*(0.068-d))
the magnetic field will be zero by right hand rule between the wires
due to wire 1 , the direction is into the page and due to the wire 2 the magnetic field is out of the page
B1 = B2
4pi*10^-7*4/(2pi*d) = 4pi*10^-7*3.4/(2pi*(0.068-d))
solving for d
d = 0.03675 m = 3.675 cm
C) charge q = 3.1*10 C = 31 C , located a distance of 2.9 cm to the right of wire I2
and moving with speedv = 46.6 m/s
F = B*q*v sin theta
F = (B1+B2)*q*v sin90
F = (mue0*I1 /(2pi*(r+0.029)) + mue0*I2 /(2pi*(0.029)))*q*v
F = ((4pi10^-7*4/(2pi(0.068+0.029)))+(4pi10^-7*3.4/(2pi(0.029))))(31*46.6) N
F = 0.04578760