A 2.0-cm length of wire centred 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.
A)1.6 nT in the negative z direction
B)1.6 nT in the positive z direction
C)2.4 nT in the negative z direction
D)None of the above

Question

Quizplus · Accepted Answer

Using the Biot-Savart law, we can find the magnetic field at point x = 5.0 m on the x-axis. The distance from the wire to the point on the x-axis is r = 5.0 m. The magnetic field is given by:

B = (μ0/4π) * I * ∫(dl × r̂)/r^2

where μ0 is the permeability of free space, I is the current, dl is an element of wire, r̂ is a unit vector in the direction of r, and the integral is taken over the length of the wire.

We can use cylindrical coordinates to simplify the calculation. Let the wire lie in the y-z plane, with dl = dŷ. Then, for a point on the x-axis at (x, 0, 0):

r = (x^2 + y^2 + z^2)^(1/2) = x
r̂ = x̂
dl × r̂ = -dŷ
r^2 = x^2

Substituting these into the formula for B, we get:

B = (μ0/4π) * I * ∫(-dŷ × x̂)/x^2

The integration is over the length of the wire, which goes from y = -1.0 cm to y = 1.0 cm. The cross product dl × r̂ is perpendicular to both dl and r̂, so it points in the -z direction. The unit vector in the -z direction is -ẑ. We can use the right-hand rule to find that dŷ × x̂ = -ẑ.

Therefore, the integral simplifies to:

B = (μ0/4π) * I * ∫ẑ/x^2 dy

Integrating over y gives:

B = (μ0/4π) * I * [ẑ/x^2 * (1.0 cm - (-1.0 cm))] = (μ0/4π) * I * ẑ/x^2 * 2.0 cm

Substituting the values, we get:

B = (4π×10^-7 Tm/A) * 20 A * (-ẑ)/0.05^2 m^2 * 0.02 m = -1.6 nT ẑ

The magnetic field is in the negative z direction, so the best choice is A).

A 20-Cm Length of Wire Centred on the Origin Carries a a 20-A