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

- A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic force that acts as centripetal force.
- Hence: \[ qvB = \frac{mv^2}{r} \]
- Rearranging for radius: \[ r = \frac{mv}{qB} \]
- Given:
- \( m = 9.1 \times 10^{-31} \, kg \)
- \( v = 4.8 \times 10^6 \, m/s \)
- \( q = 1.6 \times 10^{-19} \, C \)
- \( B = 0.182 \, T \)
- Substitute values: \[ r = \frac{(9.1 \times 10^{-31})(4.8 \times 10^6)}{(1.6 \times 10^{-19})(0.182)} \]
- Calculate numerator: \[ 9.1 \times 4.8 = 43.68 \quad \Rightarrow \quad 4.368 \times 10^{-24} \]
- Calculate denominator: \[ 1.6 \times 0.182 = 0.2912 \quad \Rightarrow \quad 2.912 \times 10^{-20} \]
- Now divide: \[ r = \frac{4.368 \times 10^{-24}}{2.912 \times 10^{-20}} \]
- Final result: \[ r = 1.5 \times 10^{-4} \, m \]
- Hence, the radius of circular path is: 1.5 × 10⁻⁴ m