Question

An electron moving with a velocity of \(4.8 \times 10^6 \, \text{ms}^{-1}\) enters a uniform magnetic field of \(0.182 \, T\) in a direction perpendicular to the field. - The radius of the circular path in which the electron moves under the influence of the magnetic field is - (Mass of electron \(= 9.1 \times 10^{-31} \, kg\) and charge of electron \(= 1.6 \times 10^{-19} \, C\))

• A. \(1.5 \times 10^{-4} \, m\)
• B. \(1.5 \times 10^{-3} \, m\)
• C. \(2.5 \times 10^{-3} \, m\)
• D. \(2.5 \times 10^{-4} \, m\)

Hint:

The radius of a charged particle’s circular motion in a magnetic field is given by: \[ r = \frac{m v}{q B} \] For electrons, always use \( m = 9.1 \times 10^{-31} \) kg and \( q = 1.6 \times 10^{-19} \) C.

Answer 1

The Correct Option is A

Step 1: Using the Formula for the Radius of Circular Motion When a charged particle moves perpendicular to a uniform magnetic field, it follows a circular path. The radius of this path is given by the formula: \[ r = \frac{m v}{q B} \] where:
- \( m = 9.1 \times 10^{-31} \) kg (mass of the electron),
- \( v = 4.8 \times 10^6 \) m/s (velocity of the electron),
- \( q = 1.6 \times 10^{-19} \) C (charge of the electron),
- \( B = 0.182 \) T (magnetic field strength). Step 2: Substituting Values \[ r = \frac{(9.1 \times 10^{-31}) (4.8 \times 10^6)}{(1.6 \times 10^{-19}) (0.182)} \] \[ r = \frac{4.368 \times 10^{-24}}{2.912 \times 10^{-20}} \] \[ r = 1.5 \times 10^{-4} \text{ m} \] Thus, the correct answer is \( \mathbf{(1)} \ 1.5 \times 10^{-4} \, m \).