Two conductors carry currents in the same direction as shown in the following fi

Question

Two conductors carry currents in the same direction as shown in the following figure. Point A is the origin of the x-axis. The positions of the two wires are -10 cm for wire(1) and +10cm for wire (2) from the origin. The position of the point B is 25 cm from the origin. The point C is at the position -25cm from the origin. Calculate the total magnetic field created by the two wires at the point A. Calculate the total magnetic field created by the two wires at the point B. Calculate the total magnetic field created by the two wires at the point C.

Explanation / Answer

magnetic field due to a infinite wire carrying current of I A, at a distance of d is given by

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

direction is given by right hand thumb rule.

hence for the wire carrying 5 A, at any distance d, magnetic field magnitude

=B1=mu*I1/(2*pi*d)

=4*pi*10^(-7)*5/(2*pi*d)=10^(-6)/d T

if the point lies to the right of the wire, direction of field will be into the plane of the paper.

if point lies to the left of the wire, direction will be out of the plane of the paper.

for the wire carrying 10 A, at any distance d, magnetic field magnitude

=B2=mu*I2/(2*pi*d)

=4*pi*10^(-7)*10/(2*pi*d)=2*10^(-6)/d T

if the point lies to the right of the wire, direction of field will be into the plane of the paper.

if point lies to the left of the wire, direction will be out of the plane of the paper.

assume that field pointing out of the plane of the paper is +ve.

part a:

point A :

lies to the right of wire 1. hence B1 is -ve.

distance from wire 1=d=10 cm=0.1 m

B1=-10^(-6)/0.1=-10^(-5) T

point A lies to the left of wire 2. hence B2 is +ve.

distance from wire 2=d=10 cm=0.1 m

B2=2*10^(-6)/0.1=2*10^(-5) T

total field=B1+B2=10^(-5) T

direction : out of the plane of the paper

part b:

point B :

lies to the right of wire 1. hence B1 is -ve.

distance from wire 1=d=35 cm=0.35 m

B1=-10^(-6)/0.35=-2.857*10^(-6) T

point B lies to the right of wire 2. hence B2 is -ve.

distance from wire 2=d=25-10 cm=15 cm=0.15 m

B2=-2*10^(-6)/0.15=-1.333*10^(-5) T

total field=B1+B2=-1.6187*10^(-5) T

direction : into the plane of the paper

part C:

point C:

lies to the left of wire 1. hence B1 is +ve.

distance from wire 1=d=15 cm=0.15 m

B1=10^(-6)/0.15=6.67*10^(-6) T

point C lies to the left of wire 2. hence B2 is +ve.

distance from wire 2=d=35 cm=0.35 m

B2=2*10^(-6)/0.35=5.71428*10^(-6) T

total field=B1+B2=1.2384*10^(-5) T

direction : out of the plane of the paper

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