A charged particle is moving on circular path with velocityv in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes

Options

A. 8 times
B. 4 times
C. 2 times
D. 16 times

Explanation

To solve the problem, we need to understand the relationship between the radius of the circular path of a charged particle moving in a magnetic field and the parameters involved: the velocity of the particle (v), the charge of the particle (q), the magnetic field strength (B), and the radius of the path (r).

### Step-by-Step Solution:

1. **Understanding the Force on the Charged Particle**:
The magnetic force acting on a charged particle moving in a magnetic field is given by:
\[ F_m = qvB \sin \theta \]
Since the velocity is perpendicular to the magnetic field, \(\theta = 90^\circ\) and \(\sin 90^\circ = 1\). Therefore, the magnetic force simplifies to:
\[ F_m = qvB \]

2. **Centripetal Force**:
The centripetal force required to keep the particle moving in a circular path 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.

3. **Equating the Forces**:
For the particle to move in a circular path, the magnetic force must equal the centripetal force:
\[ qvB = \frac{mv^2}{r} \]

4. **Solving for Radius**:
Rearranging the equation to find the radius \(r\):
\[ r = \frac{mv}{qB} \]

5. **Changing Conditions**:
According to the problem, the velocity of the charged particle is doubled:
\[ v' = 2v \]
and the strength of the magnetic field is halved:
\[ B' = \frac{B}{2} \]

6. **Finding the New Radius**:
We can now find the new radius \(r'\) using the new values:
\[ r' = \frac{mv'}{qB'} = \frac{m(2v)}{q(\frac{B}{2})} \]
Simplifying this:
\[ r' = \frac{2mv}{qB} \cdot 2 = \frac{4mv}{qB} \]

7. **Relating New Radius to Original Radius**:
From our earlier equation for \(r\):
\[ r = \frac{mv}{qB} \]
Thus, we can express \(r'\) in terms of \(r\):
\[ r' = 4 \cdot \frac{mv}{qB} = 4r \]

8. **Conclusion**:
Therefore, the new radius \(r'\) becomes:
\[ r' = 4r \]

### Final Answer:
The radius becomes \(4r\).