A long insulated wire is stretched along the x-axis. A current of 2.6 A flows in

Question

A long insulated wire is stretched along the x-axis. A current of 2.6 A flows in this wire in the positive x-direction.

Another long insulated wire is stretched along the y-axis. It carries a current of 9.2 A in the negative y-direction.

What is the magnitude of the magnetic field at the point (x=3.7cm, y=1.5cm)? The answer may be given in the units of mT or uT.

The answer should be positive if the direction of the magnetic field is up (in the positive z-direction), and negative if the direction of the magnetic field is down.

Explanation / Answer

here,

I1 = 2.6 A
I2 = -9.2 A

Dx = 3.7 cm = 0.037 m
Dy = 1.5 cm = 0.015 m

magnatic field in a current carrying wire is given by :

B = (uo*I)/(2*pi*D)
uo = magnatic permitivity

B = (4*pi*10^-7*I)/(2*pi*D)
B = (2*10^-7*I) / D -------------(1)

For Bx : eqn 1 can be written as :

Bx = (2*10^-7*2.6) / 0.037
Bx = 5.2*10^-7 / 0.037
Bx = 1.405*10^-5 T
Bx = 1.405*10^-2 mT

For Bx : eqn 1 can be written as :

Bx = (2*10^-7*(-9.2)) / 0.015
Bx = -18.4*10^-7 / 0.037
Bx = -4.972*10^-5 T
Bx = -4.972*10^-2 mT

Bnet = Sqrt(Bx^2 + By^2)
Bnet = Sqrt((1.405*10^-2)^2 + (-4.972*10^-2)^2)

Bnet = 0.05166 mT towards positive Z axis

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