Chapter 4: Q7E (page 948)

A negative charge q= -3.60 * 10-6 C is located at the origin and has velocity vS_17.50 * 104m>s2dn_1-4.90 * 104 m>s2en. At this instant what are the magnitude and direction of the magnetic field produced by this charge at the point x= 0.200 m, y= -0.300 m, z= 0?

Short Answer

Expert verified

$\vec{B} = (9.75 \times 10^{- 8} T) \hat{k} ∣$

Step by step solution

Concept.

Consider a negative charge $q = 3.60 \times 10^{- 6} C$ , which is located at the origin and has a velocity of $\vec{v} = (7.5 \times 10^{4} m / s) \hat{i} + (- 4.90 \times 10^{4} m / s) \hat{j}$ The magnetic field due to this moving electron at the nucleus, is given by,

$\vec{B} = \frac{μ_{o}}{4 π} \frac{q \vec{v} \times \vec{r}}{r^{3}}$ (1)

Step 2: Calculation.

we need to find the magnetic field produced by this charge at the point x = 0.200 m, y= -0.300 m, and z =0. So we have $\vec{r} = (0.200) \hat{i} + (- 0.300) \hat{j}$ , thus,

$\begin{matrix} \vec{v} \times \vec{r} = [(7.50 \times 10^{4} m / s) \hat{i} + (- 4.90 \times 10^{4} m / s) \hat{j}] \\ \times [(0.200 m) \hat{i} + (- 0.300 m) \hat{j}] \\ = (- 2.25 \times 10^{4} m^{2} / s) \hat{k} + (9.80 \times 10^{3} m^{2} / s) \hat{k} \\ = (- 1.27 \times 10^{4} m^{2} / s) \hat{k} \end{matrix}$

now we need to substitute into (1) but before we need to find r, as,

$\begin{matrix} r = \sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{(0.200 m)^{2} + (- 0.300 m)^{2}} = 0.3606 m \\ \vec{B} = \frac{(1.00 \times 10^{- 7} T \cdot m / A) (- 3.60 \times 10^{- 6} C) (- 1.27 \times 10^{4} m^{2} / s)}{(0.3606 m)^{3}} \hat{k} \\ = (9.75 \times 10^{- 8} T) \hat{k} \\ \vec{B} = (9.75 \times 10^{- 8} T) \hat{k} \end{matrix}$