A proton and an electron both moving with the same velocity v enter into a region of magnetic field directed perpendicular to the velocity of the particles. They will now move in cirular orbits such that

Options

A. (a) their time periods will be same
B. (b) the time period for proton electron will be higher
C. (c) the time period for electron will be higher
D. (d) their orbital radii will be same

Answer

To solve the problem, we need to analyze the motion of a proton and an electron moving in a magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Forces Acting on the Particles
When a charged particle (like a proton or an electron) moves through a magnetic field, it experiences a magnetic force that acts perpendicular to its velocity. This magnetic force provides the necessary centripetal force to keep the particle moving in a circular path.

### Step 2: Write the Expression for Centripetal Force
The centripetal force required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle, \( v \) is its velocity, and \( r \) is the radius of the circular path.

### Step 3: Write the Expression for Magnetic Force
The magnetic force acting on a charged particle moving in a magnetic field is given by:
\[ F_m = qvB \]
where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.

### Step 4: Set the Forces Equal
For circular motion, the magnetic force provides the centripetal force, so we can set these two forces equal to each other:
\[ \frac{mv^2}{r} = qvB \]

### Step 5: Solve for the Radius of the Circular Path
Rearranging the equation to find the radius \( r \):
\[ r = \frac{mv}{qB} \]

### Step 6: Relate Angular Velocity and Time Period
The angular velocity \( \omega \) of the particle is related to its linear velocity \( v \) and radius \( r \) by:
\[ \omega = \frac{v}{r} \]
Substituting the expression for \( r \):
\[ \omega = \frac{v}{\frac{mv}{qB}} = \frac{qB}{m} \]

### Step 7: Find the Time Period
The time period \( T \) of circular motion is related to angular velocity by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = \frac{2\pi m}{qB} \]

### Step 8: Compare the Time Periods of Proton and Electron
From the expression \( T = \frac{2\pi m}{qB} \), we can see that the time period \( T \) is directly proportional to the mass \( m \) of the particle. Since the mass of a proton is greater than that of an electron, we conclude:
- The time period of the proton \( T_p \) is greater than the time period of the electron \( T_e \):
\[ T_p > T_e \]

### Conclusion
Thus, the proton has a higher time period than the electron when both are moving with the same velocity in a magnetic field.