Question

Two long, parallel wires carry currents of I1 = 2.85 A and I2 = 5.45 A in the directions indicated in the figure below, where d = 19.5 cm. (Take the positive x direction to be to the right.)

(a) Find the magnitude and direction of the magnetic field at a point midway between the wires.
magnitude______  μT
direction________ ° counterclockwise from the +x axis

(b) Find the magnitude and direction of the magnetic field at point P, located d = 19.5 cm above the wire carrying the 5.45-A current.
magnitude________μT
direction________ ° counterclockwise from the +x axis

Answer 1

Magnetic field due to current carrying wire is given by:

B = u0*i/(2*pi*r)

Where direction of field is perpendicular to the direction of wire, from right hand rule.

So magnetic field due the left wire at A will be

B1 = u0*i1/(2*pi*r1) j

r1 = d/2 = 0.195/2 = 0.0975 m

i1 = 2.85 Amp

And magnetic field due the right wire at A will be

B2 = u0*i2/(2*pi*r2) (-j)

r2 = d/2 = 0.195/2 = 0.0975 m

i2 = 5.45 Amp

So net magnetic field at point A will be

Bnet = B1 + B2

Bnet = u0*i1/(2*pi*r1) j + u0*i2/(2*pi*r2) (-j)

Bnet = (u0/(2*pi))*[i1/r1 - i2/r2] j

Using known values:

Bnet = (4*pi*10^-7/(2*pi))*[2.85/0.0975 - 5.45/0.0975] j

Bnet = (-5.33*10^-6) j T

Bnet = -5.33 $\mu$ T j

Magnitude of net magnetic field = |Bnet| = 5.33 $\mu$ T

Direction = negative y-axis = 270 deg counterclockwise from the +x axis

Part B.

Part (b) magnetic field due to long current carrying wire is given by,

B = u0*I/(2*pi*r)

from figure,

B1 = u0*I1/(2*pi*r1)

here, I1 = 2.85 A

r1 = d*sqrt(2) = 0.195*sqrt(2) = 0.276 m

So, B1 = (4*pi*10^-7)*2.85/(2*pi*0.276) = 2.065*10^-6 T = 2.065 $\mu$ T

similarly for B2,

I2 = 5.45 A

r2 = d = 19.5 cm = 0.195 m

So, B2 = (4*pi*10^-7)*5.45/(2*pi*0.195) = 5.59*10^-6 T= 5.59 $\mu$ T

now, net magnetic field on point P will be,

Bpx = B1*cos(90 + $\theta$ ) + B2*cos(180 deg)

from figure, $\theta$ = arctan(d/d) = 45 deg

So, Bpx = 2.065*cos(135 deg) + 5.59*cos(180 deg)

Bpx = -7.05 $\mu$ T

also, Bpy = B1*sin(90+ $\theta$ ) = 2.065*sin(135 deg)

Bpy = 1.46 $\mu$ T

So, net magnetic field = Bp = sqrt(Bpx^2 + Bpy^2)

Bp = sqrt((-7.05)^2 + 1.46^2)

Bp = 7.2 $\mu$ T

direction = $\phi$ = arctan(Bpy/Bpx) = arctan(1.46/(-7.05)) = -11.7 deg

Or, $\phi$ = 180 - 11.7 = 168.3 deg counterclockwise from +x-axis

Let me know if you've any query.