Three long, parallel, straight wires pass through the vertices of an equilateral

Question

Three long, parallel, straight wires pass through the vertices of an equilateral triangle that has sides L = 7.00 cm, as shown in the figure below. A a dot indicates that the direction of the current is out of the screen and a cross indicates that the direction of the current is into the screen.

Each current is 14.0 A.

(a) Find the magnetic field at the location of the upper wire due to the currents in the two lower wires.

(b) Find the force per unit length on the upper wire.

Explanation / Answer

part A :

magnetic field due to a straight current carrying wire is given by B = Uo I/2piR

here Due to left corner Current B1 = uoI/2piR

here R = distance between L = 0.07 m

so

B1 = (4pi *10^-7 * 14)/(2* 3.14* 0.07)

B1 = 4*10^-5 Tesla , Clearly by RH rule this Directed in CW direction at dot.

since all the magnitudes of other wire are also same

B2 also will be = 4*10^-5 Tesla, as this is directed in CW Direction i.e towards right

so BNet at Dot = B1 + B2 = 8*10^-5 Tesla

-------------------

Force = I L B

Force per length = F/L = Bnet * I

F/L = 8*10^-5 * 14

F/L = 1.12*10^-3 N towards right

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