1. The radius of a coil of wire with N turns is r = 0.28 m. A clockwise current

Question

1. The radius of a coil of wire with N turns is r = 0.28 m. A clockwise current of Icoil = 1.0 A flows in the coil, as shown. A long, straight wire carrying a current Iwire = 29 A toward the left is located 0.04 m from the edge of the coil. The magnetic field at the center of the coil is zero tesla.

a) Determine the strength and direction of the magnetic field due to the straight wire at a distance equal to the center of the loop.

b) The magnetic field of the loop cancels the magnetic field of the wire. Explain why this is so.

c) Determine N, the number of turns of the coil.

Explanation / Answer

a)

Magnetic field due to long current carrying wire = B = u0I/2pieR = 4 pie x 10^-7 x 29/ 2 pie x 0.32

Therefore B = 181.25 x 10-7 T

Direction will be in z direction that is outside the plane of paper.

b)

This is because the magnetic fields produced by the coil and the wire are in opposite direction at the center of loop so they can cancel each other and the net magnetic field at the center becomes 0.

c)

Magnetic field due to coil = uo x N x I/ 2 r in the -z direction given by right hand thumb rule.

And net magnetic field at center = 0

Therefore, Bloop = Bwire

u0 N I1 / 2 r = u0 I / 2 pie R => N= 1/ pie x r/R x I /I1= 0.278 x 29/1 = 8

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