Two parallel wires 1 and 2 carry current I and 2I, respectively, flowing in the

Question

Two parallel wires 1 and 2 carry current I and 2I, respectively, flowing in the same direction as shown in the figure, where I = 0.18 A. The magnitude of the magnetic field due to current in wire 1 at a location on wire 2 is 3.2 x 10-6 T.

What is the separation of the wires, d?

A) 1.1 x 10^-2 m

B) 2.3 x 10^-2 m

C) 3.4 x 10^-2 m

D) 30 m

E) 89 m

What is the magnetic force per unit length on wire 2? Are the wires attracted to each other or repelled each other?

A) 5.8 x 10^-7 F/m, attract

B) 1.2 x 10^-6 F/m, attract

C) 1.7 x 10^-6 F/m, attract

D) 1.2 x 10^-6 F/m, repel

E) 1.7 x 10^-6 F/m, repel

Explanation / Answer

part 1.

magnetic field due to a current carrying wire is given by

B=mu*I/(2*pi*d)

where mu=magnetic permeability=4*pi*10^(-7)

I=current in the wire.

d=distance of the point from the wire.

now, given that B=3.2*10^(-6) T

I=0.18 A

then d=mu*I/(2*pi*B)=0.01125 m=1.125 cm

hence option A is correct.

part 2:

as both wires carrying current in same direction, force will be attractive in nature.

force per unit length=mu*I1*I2/(2*pi*d)

=mu*I*2*I/(2*pi*d)=1.152*10^(-6) N/m

hence option B is correct.

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