One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?
A) 4.0 μT
B) 5.0 μT
C) 3.0 μT
D) 7.0 μT
E) 1.0 μT

Question

Quizplus · Accepted Answer

We will use the Biot-Savart law to find the total magnetic field at the point (0, 2, 0) m on the y axis. Since the current in the first wire is parallel to the y axis, it will not contribute to the magnetic field at this point.

Let's consider the contribution to the magnetic field due to the second wire. We will use the right hand rule to find the direction of the magnetic field.

At point (0, 2, 0) m, the vector from the wire to the point is in the negative y direction (i.e. (0, -2, 0) m). The current is in the positive z direction. Curl your fingers from the current direction (z axis) towards the vector pointing towards the point (y axis), and your thumb will point in the negative x direction. So the magnetic field due to the second wire at point (0, 2, 0) m is in the negative x direction.

We can use the Biot-Savart law to find the magnitude of the magnetic field due to the second wire:

$\mathrm{d}\vec{B} = \frac{\mu_0}{4\pi}\frac{I\mathrm{d}\vec{l}	imes\vec{r}}{r^2}$

where $\mathrm{d}\vec{l}$ is a small segment of the wire, $I$ is the current, $\vec{r}$ is the vector from the wire segment to the point where we want to find the magnetic field, and $r$ is the magnitude of that vector.

Choosing a small segment of the wire at position $\vec{l} = (0, y, 0) $ m, where $y$ ranges from 0 to 4 m, the magnitude of $\mathrm{d}\vec{l}$ is $\mathrm{d}l = \mathrm{d}y$. We can write $\mathrm{d}\vec{l} = \mathrm{d}y\,\hat{j}$. The vector from the wire segment to the point where we want to find the magnetic field is $\vec{r} = (-y, 2, 0)$ m. The magnitude of this vector is $r = \sqrt{y^2 + 4}$. Since the current in the second wire is constant, we can take it out of the integral:

$\vec{B} = \frac{\mu_0}{4\pi}I\int_0^4 \frac{\mathrm{d}\vec{l}	imes\vec{r}}{r^2}$

$=\frac{\mu_0}{4\pi}I\int_0^4 \frac{\mathrm{d}y\,\hat{j}	imes(-y\hat{i} + 2\hat{j})}{(y^2+4)^{3/2}}$

$=-\frac{\mu_0}{4\pi}I\int_0^4 \frac{2\mathrm{d}y\,\hat{k}}{(y^2+4)^{3/2}}$

$=-\frac{\mu_0}{2\pi}I\left[\frac{1}{2(y^2+4)^{1/2}}ight]_0^4\hat{k}$

$=\frac{\mu_0}{4\pi}\left(\frac{40}{2	imes 2^{3/2}}ight)\hat{i}$

$=5.0\,\mu\mathrm{T}\,\hat{i}$

So the magnitude of the total magnetic field at the point (0, 2, 0) m on the y axis is 5.0 μT. Therefore, the best choice is B.

One Long Wire Carries a Current of 30 a Along