Two straight current-carrying conductors A and B are positioned perpendicular to
ID: 1437006 • Letter: T
Question
Two straight current-carrying conductors A and B are positioned perpendicular to the page along the x-axis. The distance between them is 30.0 cm. I_A = 3.50 A, I_B = 2.00 A. Point P_1 is located between the two currents at a distance of 10.0cm from I_A. What is the direction and magnitude of the net magnetic field at P_1? The net magnetic field due to the two currents is zero at a point P_2 on the x axis (at a finite distance from the currents). Is P_2 located to the left of A, between the currents, or to the right of B? What is the distance of P_2 from B? If another straight current I_c (also perpendicular to the page) were to run through P2 and if the net magnetic force due to I_A and I_c on I_B were to be zero, what would be the magnitude and direction of I_c?Explanation / Answer
(a)
Net magnetic Field at point p1, B = (uo*IA)/(2*pi*ra) + (uo*IB)/(2*pi*rb)
Bnet = (2*10^-7*3.5)/0.1 + (2*10^-7*2.0)/0.2
Bnet = 9 * 10^-6 T
Direction = Towards -ve Y axis.
(b)
As Ia > Ib ,
P2 will be located to the Right of B.
(c)
Let the distance of P2 from B = x
So,Bnet at point x = 0
(uo*IA)/(2*pi*(r+x)) = (uo*IB)/(2*pi*x)
IA/0.3+x = IB/x
3.5/(0.3+x) = 2.0/x
x = 0.4 m
Distance of point p2 from B , = 40 cm
(d)
Net Magnetic Force due to IA & IC on IB = 0
So,
uo*IA*IB/(2*pi*0.3) = uo*IC*IB/(3*pi*0.4)
IA/0.3 = IC/0.4
3.5/0.3 = IC/0.4
IC = 4.67 A
Direction of Current, Into the page.
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