If the value of magnetic field is reduced to half and velocity of charged particle is doubled then the radius of path of charged particle will be:

Options

A. 2 times
B. 8 times
C. 3 times
D. 4 times

Explanation

To solve the problem, we need to understand the relationship between the radius of the path of a charged particle moving in a magnetic field and the parameters involved. The formula for the radius \( r \) of the circular path of a charged particle in a magnetic field is given by:

\[ r = \frac{mv}{qB} \]

where:
- \( m \) is the mass of the charged particle,
- \( v \) is the velocity of the charged particle,
- \( q \) is the charge of the particle,
- \( B \) is the magnetic field strength.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- Let the initial magnetic field be \( B \).
- Let the initial velocity be \( v \).
- The initial radius \( r \) can be expressed as:
\[ r = \frac{mv}{qB} \]

2. **Apply the changes as per the problem statement:**
- The magnetic field is reduced to half, so the new magnetic field \( B' \) is:
\[ B' = \frac{B}{2} \]
- The velocity of the charged particle is doubled, so the new velocity \( v' \) is:
\[ v' = 2v \]

3. **Substitute the new values into the radius formula:**
- The new radius \( r' \) can be calculated using the modified values:
\[ r' = \frac{mv'}{qB'} \]
- Substitute \( v' \) and \( B' \):
\[ r' = \frac{m(2v)}{q\left(\frac{B}{2}\right)} \]

4. **Simplify the expression:**
- This simplifies to:
\[ r' = \frac{2mv}{q\left(\frac{B}{2}\right)} = \frac{2mv}{\frac{qB}{2}} = \frac{2mv \cdot 2}{qB} = \frac{4mv}{qB} \]

5. **Relate the new radius to the initial radius:**
- Recall the initial radius \( r = \frac{mv}{qB} \):
\[ r' = 4 \left(\frac{mv}{qB}\right) = 4r \]

### Conclusion:
Thus, the new radius \( r' \) of the path of the charged particle, after the changes in magnetic field and velocity, is:

\[ \boxed{4r} \]