If a proton enters perpendicularly a magnetic field with velocity v, then time period of revolution is T. If proton enters with velocity 2v, then time period will be

Options

A. T
B. 2T
C. 3T
D. 4T

Answer

To solve the problem, we need to analyze the motion of a proton in a magnetic field when it enters perpendicularly with different velocities.

### Step-by-step Solution:

1. **Understanding the Motion in a Magnetic Field**:
When a charged particle like a proton enters a magnetic field perpendicularly, it experiences a magnetic force that causes it to move in a circular path. The magnetic force acting on the proton is given by the Lorentz force formula:
\[ F = q(v \times B) \]
where \( q \) is the charge of the proton, \( v \) is its velocity, and \( B \) is the magnetic field strength.

2. **Equating Forces**:
For circular motion, the magnetic force provides the necessary centripetal force. Therefore, we can equate the magnetic force to the centripetal force:
\[ qvB = \frac{mv^2}{r} \]
where \( m \) is the mass of the proton and \( r \) is the radius of the circular path.

3. **Finding the Radius of Circular Motion**:
Rearranging the equation gives us:
\[ r = \frac{mv}{qB} \]

4. **Calculating the Time Period**:
The time period \( T \) of the circular motion is given by the formula:
\[ T = \frac{2\pi r}{v} \]
Substituting the expression for \( r \):
\[ T = \frac{2\pi \left(\frac{mv}{qB}\right)}{v} \]
Simplifying this, we find:
\[ T = \frac{2\pi m}{qB} \]
This shows that the time period \( T \) does not depend on the velocity \( v \).

5. **Analyzing the New Velocity**:
If the proton enters the magnetic field with a velocity \( 2v \), we can repeat the same analysis. The radius of the circular path would now be:
\[ r' = \frac{m(2v)}{qB} = 2 \left(\frac{mv}{qB}\right) = 2r \]
However, when we calculate the time period with the new radius:
\[ T' = \frac{2\pi r'}{2v} = \frac{2\pi (2r)}{2v} = \frac{2\pi r}{v} = T \]

6. **Conclusion**:
Therefore, the time period of revolution remains the same, regardless of the change in velocity. Thus, if the proton enters with velocity \( 2v \), the time period will still be \( T \).

### Final Answer:
The time period will be \( T \).