Biot-Savart law: A point charge Q moves on the x-axis in the positive direction with a speed of A point P is on the y-axis at The magnetic field produced at the point P, as the charge moves through the origin, is equal to What is the charge Q? (μ₀ = 4π × 10^-7 T ∙ m/A)

Question

Quizplus · Accepted Answer

Using the Biot-Savart law, we have:

$$\vec{B} = \frac{\mu_0}{4\pi} \frac{q\vec{v}	imes\vec{r}}{r^3}$$

where $\mu_0$ is the permeability of free space, $\vec{v}$ is the velocity of the point charge, $\vec{r}$ is the vector from the point charge to the point of interest, and $q$ is the charge of the point charge.

In this case, the charge is moving on the positive x-axis, so $\vec{v}=v_x\hat{x}$. The point of interest is on the positive y-axis, so $\vec{r}=r\hat{y}$ and $r= \sqrt{(0.1)^2+(0.2)^2}=0.22\ m$. Plugging in the given values for $\vec{B}$ and $\mu_0$, we get:

$$\frac{0.06 q v_x}{(0.22)^2}=(2.1)	imes 10^{-6}\ T$$

where we have converted the given value of $\vec{B}$ to its magnitude. Solving for $q$, we get:

$$q = \frac{(0.22)^2(2.1)	imes 10^{-6}}{0.06v_x}$$

Plugging in the given value of $v_x$, we get:

$$q = -53	imes 10^{-6}\ C$$

So the answer is choice A), with a charge of $-53\ \mu C$.

Biot-Savart Law: a Point Charge Q Moves on the X-Axis