Suppose that a proton travelling in vacuum with velocity `V_(1)` at right angles to a uniform magnetic field experiences twice the force that an `alpha`-particle experiences when it is travelling along into same path with velocity `V_(2)`. The ratio `(V_(1))/(V_(2))` is

Options

A. 0.5
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to analyze the forces acting on the proton and the alpha particle in a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Forces**:
The force experienced by a charged particle moving in a magnetic field is given by the formula:
\[ F = QvB \sin \theta \]
where \( F \) is the magnetic force, \( Q \) is the charge of the particle, \( v \) is the velocity of the particle, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. In this case, since the proton is moving at right angles to the magnetic field, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \).

2. **Force on the Proton**:
For the proton, we have:
\[ F_1 = Q_1 V_1 B \]
where \( Q_1 \) is the charge of the proton, \( V_1 \) is its velocity, and \( B \) is the magnetic field strength.

3. **Force on the Alpha Particle**:
An alpha particle consists of 2 protons and 2 neutrons, thus it has a charge of:
\[ Q_2 = 2Q_1 \]
For the alpha particle moving along the same path, we have:
\[ F_2 = Q_2 V_2 B = 2Q_1 V_2 B \]

4. **Given Condition**:
According to the problem, the force on the proton is twice that on the alpha particle:
\[ F_1 = 2F_2 \]
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ Q_1 V_1 B = 2(2Q_1 V_2 B) \]
Simplifying this gives:
\[ Q_1 V_1 B = 4 Q_1 V_2 B \]

5. **Canceling Common Terms**:
Since \( Q_1 \) and \( B \) are common on both sides and non-zero, we can cancel them:
\[ V_1 = 4 V_2 \]

6. **Finding the Ratio**:
To find the ratio \( \frac{V_1}{V_2} \):
\[ \frac{V_1}{V_2} = 4 \]

### Final Answer:
The ratio \( \frac{V_1}{V_2} \) is \( 4 \).

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