Question

In an experiment, electrons are accelerated, from rest, by applying a voltage of $500\mathrm{V}$. Calculate the radius of the path if a magnetic field $100\mathrm{mT}$ is applied.

[Charge of the electron$=1.6\times {10}^{-19}\mathrm{C}$, Mass of the electron $=9.1\times {10}^{-31}\mathrm{kg}$]

• A. $7.5\times {10}^{-4}\mathrm{m}$
• B. $7.5\times {10}^{-3}\mathrm{m}$
• C. $7.5\mathrm{m}$
• D. $7.5\times {10}^{-2}\mathrm{m}$

Answer 1

The correct option is A

$7.5\times {10}^{-4}\mathrm{m}$

Explanation for correct answer:

Option (a) $7.5\times {10}^{-4}\mathrm{m}$

Step 1: Given data

Voltage $=500\mathrm{V}$

Magnetic field strength $=100\mathrm{mT}$

Charge of the electron$=1.6\times {10}^{-19}\mathrm{C}$,

Mass of the electron $=9.1\times {10}^{-31}\mathrm{kg}$

Step 2: Formula used

To find the required radius, we will use the formula given below:

$$\begin{gathered} \mathrm{r}=\frac{\sqrt{2\mathrm{mk}}}{\mathrm{eB}} \\ \mathrm{r}=\frac{\sqrt{2\mathrm{me}\Delta \mathrm{v}}}{\mathrm{eB}} \end{gathered}$$

Where, $\mathrm{m}$ is the mass of electron, $\mathrm{v}$is voltage, $\mathrm{e}$ is charge on electron and $\mathrm{B}$ is magnetic field strength in Tesla.

Step 3: Find the value of radius

Now, we will substitute the given values in the formula:

$\mathrm{r}=\frac{\sqrt{2\mathrm{me}\Delta \mathrm{v}}}{\mathrm{eB}}$

$\mathrm{r}=\frac{\sqrt{\frac{2\mathrm{m}}{\mathrm{e}}·\Delta \mathrm{v}}}{\mathrm{B}}=\frac{\sqrt{\frac{2\times 9.1\times {10}^{-31}}{1.6\times {10}^{-19}\left(500\right)}}}{100\times {10}^{-3}}$

$\mathrm{r}=\frac{\sqrt{\frac{9.1}{0.16}\times {10}^{-10}}}{{10}^{-1}}=\frac{3}{4}\times {10}^{-4}=7.5\times {10}^{-4}\mathrm{m}$

Hence, correct answer is (a).