Question

A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0 m on the x-axis.

Answer 1

Answer:

The magnetic field is 1.6 nT in z-direction

Explanation:

Given;

Length of wire, L = 2.0-cm

Current in the wire, I = 20-A

point from the axis, r = 5.0 m

Strength of magnetic field can be calculated by applying Biot-Savart equation;

$B = \frac{\mu I}{2\pi r} (\frac{L}{\sqrt{4r^2 } +L} )$

where;

μ is constant given as 4π x 10⁻⁷ T.m/A

Substitute the given values in the equation above and calculate strength of magnetic field.

Since current is moving in y-direction and magnetic field will be caculated from a point 5m in the x-direction, based on vector law, the resultant direction will be z.

$B = \frac{\mu I}{2\pi r} (\frac{L}{\sqrt{4r^2 } +L} )\\\\B = \frac{4\pi *10^{-7} *20}{2\pi *5} (\frac{0.02}{\sqrt{4(5)^2 } +0.02} )\\\\B = 8*10^{-7} *1.9998 *10^{-3}\\\\B = 1.6*10^{-9} \ T\\\\B = 1.6 \ nT$

Therefore, the magnetic field is 1.6 nT in z-direction

Answer 2

Answer: field strength = 8x10^-8T

Explanation:

At a distance r from a long current carrying conductor of current I,

B = uL/2¶r

Where u = 4¶x10^-7 H/m = permeability of free space.

r = 5m

I = 20A

B = (4¶x10^-7 x 20)/2¶x5

B = 8x10^-8T