Problem

An electron is projected into a uniform magnetic field of $10 \, \mathrm{mT}$ and moves in a circular path with a radius of $6 \, \mathrm{cm}$. Determine the time period of the electron’s motion.

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Data

• Radius of the circular path: $r = 6 \, \mathrm{cm} = 6 \times 10^{-2} \, \mathrm{m}$,
• Magnetic field: $B = 10 \, \mathrm{mT} = 10 \times 10^{-3} \, \mathrm{T}$,
• Charge of the electron: $q = 1.6 \times 10^{-19} \, \mathrm{C}$,
• Mass of the electron: $m = 9.11 \times 10^{-31} \, \mathrm{kg}$.

To find:

• Time period of motion: $T = ?$.

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Prerequisite Concepts

The time period of an electron moving in a circular path in a magnetic field is given by:

$$T = \frac{2 \pi m}{q B}$$

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Solution

1. Substitute the known values into the formula:

$$T = \frac{2 \pi \cdot 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \cdot 10 \times 10^{-3}}$$

2. Simplify the denominator:

$$q B = 1.6 \times 10^{-19} \cdot 10 \times 10^{-3} = 1.6 \times 10^{-21}$$

3. Simplify the numerator:

$$2 \pi m = 2 \cdot 3.14 \cdot 9.11 \times 10^{-31} = 5.72 \times 10^{-30}$$

4. Divide the values:

$$T = \frac{5.72 \times 10^{-30}}{1.6 \times 10^{-21}} = 3.575 \times 10^{-9} \, \mathrm{s}.$$

5. Express in nanoseconds:

$$T = 3.6 \, \mathrm{ns}.$$

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Answer

The time period of the electron’s motion is:

$$T = 3.6 \, \mathrm{ns}.$$