Question

Two charged particles of same mass but charges in ratio 1:4 enter a uniform perpendicular magnetic field. The ratio of their time periods is:

• A. 1:4
• B. 1:8
• C. 8:1
• D. 4:1
• E. 2:1

Hint:

Time period in a magnetic field is independent of velocity and radius.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When a charged particle enters a uniform magnetic field perpendicular to its velocity, it undergoes uniform circular motion. The magnetic force provides the necessary centripetal force. We need to find the formula for the time period of this motion and see how it depends on charge.
Step 2: Key Formula or Approach:
1. Magnetic Force: \( F_m = qvB \), where q is charge, v is velocity, B is magnetic field. 2. Centripetal Force: \( F_c = \frac{mv^2}{r} \), where m is mass, v is velocity, r is radius of the circular path. 3. Equating the forces: \( qvB = \frac{mv^2}{r} \). 4. Time Period (T): The time taken to complete one circle is the circumference divided by the speed: \( T = \frac{2\pi r}{v} \). Step 3: Detailed Explanation:
Let's derive the formula for the time period. From the force balance equation: \[ qvB = \frac{mv^2}{r} \] We can solve for the radius r: \[ r = \frac{mv}{qB} \] Now, substitute this expression for r into the time period formula: \[ T = \frac{2\pi}{v} \left( \frac{mv}{qB} \right) \] The velocity 'v' cancels out: \[ T = \frac{2\pi m}{qB} \] This is the formula for the time period of a charged particle in a uniform magnetic field. We can see that the time period is independent of the particle's velocity and the radius of its path. It depends on the mass, charge, and the magnetic field strength. Now, let's apply this to our problem. We are given: - Same mass: \( m_1 = m_2 = m \) - Same magnetic field: \( B_1 = B_2 = B \) - Ratio of charges: \( q_1 : q_2 = 1 : 4 \). Let \( q_1 = q \) and \( q_2 = 4q \). We need to find the ratio of their time periods, \( T_1 : T_2 \). \[ T_1 = \frac{2\pi m}{q_1 B} = \frac{2\pi m}{qB} \] \[ T_2 = \frac{2\pi m}{q_2 B} = \frac{2\pi m}{(4q)B} \] Now, find the ratio \( \frac{T_1}{T_2} \): \[ \frac{T_1}{T_2} = \frac{\frac{2\pi m}{qB}}{\frac{2\pi m}{4qB}} = \frac{2\pi m}{qB} \times \frac{4qB}{2\pi m} = \frac{4}{1} \] So, the ratio \( T_1 : T_2 \) is 4:1. Step 4: Final Answer:
The ratio of their time period is 4:1.