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 solve the problem, we need to find the relationship between the magnetic force acting on a charged particle and the radius of its circular path. The steps to derive the value of \( n \) are as follows:

### Step-by-Step Solution:

1. **Understand the Magnetic Force**: The magnetic force \( F \) acting on a charged particle moving in a magnetic field is given by the equation:
\[ F = Q \cdot (V \times B) \]
where \( Q \) is the charge of the particle, \( V \) is its velocity, and \( B \) is the magnetic field strength. Since the particle is projected perpendicular to the magnetic field, the angle \( \theta \) between \( V \) and \( B \) is 90 degrees.

2. **Calculate the Magnitude of the Force**: The magnitude of the force when \( \theta = 90^\circ \) is:
\[ F = Q \cdot V \cdot B \cdot \sin(90^\circ) = Q \cdot V \cdot B \]

3. **Relate Velocity to Radius**: The charged particle moves in a circular path due to the magnetic force acting as the centripetal force. The radius \( r \) of the circular path is given by the formula:
\[ r = \frac{mv}{Q B} \]
where \( m \) is the mass of the particle.

4. **Express Velocity in Terms of Radius**: Rearranging the formula for radius gives us:
\[ v = \frac{Q B r}{m} \]

5. **Substitute Velocity Back into the Force Equation**: Now we substitute this expression for \( v \) back into the equation for magnetic force:
\[ F = Q \cdot V \cdot B = Q \cdot \left(\frac{Q B r}{m}\right) \cdot B \]
Simplifying this gives:
\[ F = \frac{Q^2 B^2 r}{m} \]

6. **Identify the Relationship**: From the equation \( F = \frac{Q^2 B^2}{m} \cdot r \), we can see that the magnetic force \( F \) is directly proportional to the radius \( r \):
\[ F \propto r \]

7. **Determine the Value of \( n \)**: Since the force is proportional to \( r^1 \), we conclude that:
\[ n = 1 \]

### Final Answer:
The value of \( n \) is \( 1 \).