Question

A charged particle gains a speed of \(10^6 \, \text{ms}^{-1}\) when accelerated from rest through a potential difference of 10 kV. It enters a region of magnetic field of 0.4 T such that \( \vec{v} \perp \vec{B} \). The radius of the circular path described by it is:

• A. 2.5 cm
• B. 5 cm
• C. 8 cm
• D. 10 cm

Hint:

For a charged particle moving in a magnetic field perpendicular to the field, the radius of the circular path can be found using the formula \( r = \frac{mv}{qB} \), where \(v\) is the velocity, \(q\) is the charge, and \(B\) is the magnetic field strength.

Answer 1

The Correct Option is B

When a charged particle moves in a magnetic field at a perpendicular angle to the field, it follows a circular path. The radius \(r\) of this circular path is given by the formula: \[ r = \frac{mv}{qB} \] Where: 
- \(m\) is the mass of the particle, 
- \(v\) is the speed of the particle, 
- \(q\) is the charge of the particle, 
- \(B\) is the magnetic field strength. We know the following: 
- The particle gains a speed \(v = 10^6 \, \text{ms}^{
-1}\) after being accelerated through a potential difference \(V = 10 \, \text{kV} = 10^4 \, \text{V}\). 
- The magnetic field \(B = 0.4 \, \text{T}\). 
- The energy gained by the particle is equal to the work done by the electric field, which can be expressed as: \[ \frac{1}{2} mv^2 = qV \] From this equation, we can solve for \(m\) (mass of the particle) in terms of \(q\) (charge of the particle) and \(v\): \[ m = \frac{2qV}{v^2} \] Substituting this expression for \(m\) into the formula for \(r\): \[ r = \frac{\left( \frac{2qV}{v^2} \right) v}{qB} = \frac{2V}{vB} \] Now, substitute the given values: - \(V = 10^4 \, \text{V}\), - \(v = 10^6 \, \text{ms}^{-1}\), - \(B = 0.4 \, \text{T}\): \[ r = \frac{2 \times 10^4}{10^6 \times 0.4} = \frac{2 \times 10^4}{4 \times 10^5} = 0.05 \, \text{m} = 5 \, \text{cm} \] Thus, the radius of the circular path described by the particle is 5 cm. Therefore, the correct answer is option (B).