An electron and a proton with equal momentum enter perpendicularly into a uniform magnetic field, then

Options

A. (a) the path of proton shall be more curved then that of electron
B. (b) the path of proton shall be less curved then that of electron
C. (c) Both are equally curved
D. (d) path of both will be straight line

Answer

To solve the problem, we need to analyze the motion of an electron and a proton entering a magnetic field perpendicularly with equal momentum. We'll derive the relationship for the radius of their circular paths in the magnetic field and determine the implications of their equal momentum.

### Step-by-Step Solution:

1. **Understand the Motion in a Magnetic Field**:
When a charged particle moves in a magnetic field perpendicular to its velocity, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path.

2. **Formula for Radius of Circular Motion**:
The radius \( r \) of the circular path of a charged particle in a magnetic field is given by the formula:
\[ r = \frac{mv}{Bq} \]
where:
- \( m \) = mass of the particle,
- \( v \) = velocity of the particle,
- \( B \) = magnetic field strength,
- \( q \) = charge of the particle.

3. **Relate Momentum to Velocity**:
The momentum \( p \) of a particle is defined as:
\[ p = mv \]
Therefore, we can express velocity \( v \) in terms of momentum:
\[ v = \frac{p}{m} \]

4. **Substitute Velocity into the Radius Formula**:
Substituting \( v \) into the radius formula gives:
\[ r = \frac{m \left(\frac{p}{m}\right)}{Bq} = \frac{p}{Bq} \]

5. **Comparison of Electron and Proton**:
Since the question states that both the electron and the proton have equal momentum, we can denote their momenta as \( p_e = p_p \). The charge of the electron (\( q_e \)) is equal in magnitude but opposite in sign to the charge of the proton (\( q_p \)). Thus, we have:
- For the electron:
\[ r_e = \frac{p}{Bq_e} \]
- For the proton:
\[ r_p = \frac{p}{Bq_p} \]

6. **Evaluate the Radii**:
Since \( q_e = -e \) and \( q_p = +e \) (where \( e \) is the elementary charge), we can see that:
\[ r_e = \frac{p}{B(-e)} \quad \text{and} \quad r_p = \frac{p}{Be} \]
The absolute values of the radii are:
\[ |r_e| = \frac{p}{Be} \quad \text{and} \quad |r_p| = \frac{p}{Be} \]
This shows that the magnitudes of the radii for both particles are equal.

7. **Conclusion**:
Since both particles have equal momentum and the same magnitude of charge (but opposite signs), they will have the same radius of curvature in the magnetic field. Therefore, both the electron and the proton will follow circular paths of equal radius.

### Final Answer:
Both the electron and the proton will be equally curved in their paths in the magnetic field.

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