Magnetic force acting on the charged particle projected perpendicular to magnetic field is proportional to `r^(n)`, where r is radius of circular path. Find n

Answer

To find the value of \( n \) in the relationship where the magnetic force acting on a charged particle projected perpendicular to a magnetic field is proportional to \( r^n \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Forces**:
The magnetic force \( F_m \) acting on a charged particle moving with velocity \( v \) in a magnetic field \( B \) is given by:
\[ F_m = qvB \]
where \( q \) is the charge of the particle.

2. **Centripetal Force Requirement**:
When a charged particle moves in a circular path due to the magnetic force, this magnetic force acts as the centripetal force. The centripetal force \( F_c \) required to keep the particle in circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle and \( r \) is the radius of the circular path.

3. **Setting the Forces Equal**:
Since the magnetic force provides the centripetal force, we can set these two forces equal:
\[ qvB = \frac{mv^2}{r} \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ mv^2 = qvBr \]
Dividing both sides by \( v \) (assuming \( v \neq 0 \)):
\[ mv = qBr \]

5. **Solving for Velocity**:
From the equation \( mv = qBr \), we can express the velocity \( v \) as:
\[ v = \frac{qBr}{m} \]

6. **Substituting Velocity Back**:
Substitute \( v \) back into the expression for the magnetic force:
\[ F_m = qvB = q\left(\frac{qBr}{m}\right)B = \frac{q^2B^2r}{m} \]

7. **Identifying the Proportionality**:
From the equation \( F_m = \frac{q^2B^2r}{m} \), we can see that the magnetic force \( F_m \) is proportional to \( r \):
\[ F_m \propto r \]
This implies that \( n = 1 \).

### Conclusion:
Thus, the value of \( n \) is:
\[ \boxed{1} \]