Question

The radius of the path of an electron moving at a speed of $ 3.2 \times 10^7 $ ms$ ^{-1} $ in a magnetic field of $ 6 \times 10^{-4} $ T perpendicular to it is (mass of electron is $ 9 \times 10^{-31} $ kg and charge of electron is $ 1.6 \times 10^{-19} $ C)

• A. 22.4 cm
• B. 13 cm
• C. 30 cm
• D. 39 cm

Hint:

The magnetic force provides the centripetal force for circular motion.

Answer 1

The Correct Option is C

Step 1: Identify given values.
$ v = 3.2 \times 10^7 \, \text{ms}^{-1} $
$ B = 6 \times 10^{-4} \, \text{T} $
$ m = 9 \times 10^{-31} \, \text{kg} $
$ q = 1.6 \times 10^{-19} \, \text{C} $
Step 2: Equate magnetic force to centripetal force. $ |q| v B = \frac{mv^2}{r} $ Step 3: Solve for radius $ r $.
$ r = \frac{mv}{|q| B} $ Step 4: Substitute values.
$ r = \frac{(9 \times 10^{-31})(3.2 \times 10^7)}{(1.6 \times 10^{-19})(6 \times 10^{-4})} \, \text{m} $
$ r = \frac{28.8 \times 10^{-24}}{9.6 \times 10^{-23}} \, \text{m} = 0.3 \, \text{m} $ Step 5: Convert to cm.
$ r = 0.3 \times 100 = 30 \, \text{cm} $
Step 6: Conclusion.
The radius is 30 cm.