A small particle of mass m and charge Q is dropped in uniform horizontal magnetic field B. The maximum vertical displacement of particle is given by `h=(nm^2g)/(2Q^2B^2)` . Find the value n.

Answer

To solve the problem, we need to analyze the motion of a charged particle in a magnetic field and derive the maximum vertical displacement \( h \) in terms of the given parameters.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- A particle of mass \( m \) and charge \( Q \) is dropped in a uniform horizontal magnetic field \( B \).
- The gravitational force acting on the particle is \( F_g = mg \) (downward).
- The magnetic force acting on the particle is given by \( F_m = Q \mathbf{v} \times \mathbf{B} \). Since the magnetic field is horizontal and the particle is dropped, the magnetic force will act perpendicular to the velocity of the particle.

2. **Motion in the Magnetic Field**:
- As the particle falls under gravity, it also experiences a magnetic force that causes it to move in a circular path due to the Lorentz force.
- The magnetic force can be expressed as \( F_m = QvB \) when the angle between the velocity and the magnetic field is 90 degrees.

3. **Components of Motion**:
- The particle has a vertical motion due to gravity and a horizontal motion due to the magnetic force.
- The vertical motion can be described by the equation of motion under gravity, while the horizontal motion is circular due to the magnetic force.

4. **Finding the Relationship**:
- The maximum vertical displacement \( h \) can be derived from the balance of forces. The magnetic force provides the centripetal force required for circular motion.
- The centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( r \) is the radius of the circular path.
- Setting the magnetic force equal to the centripetal force gives us \( QvB = \frac{mv^2}{r} \).

5. **Using Energy Conservation**:
- The particle starts from rest and falls under gravity. The potential energy lost is converted into kinetic energy.
- The potential energy at height \( h \) is \( mgh \) and the kinetic energy when it reaches the bottom is \( \frac{1}{2} mv^2 \).
- Setting these equal gives us \( mgh = \frac{1}{2} mv^2 \) which simplifies to \( v^2 = 2gh \).

6. **Substituting for \( v \)**:
- Substitute \( v^2 \) into the centripetal force equation:
\[ QvB = \frac{m(2gh)}{r} \]
- Rearranging gives us a relationship involving \( h \).

7. **Final Expression for \( h \)**:
- After manipulating the equations, we arrive at:
\[ h = \frac{nm^2g}{2Q^2B^2} \]
- Comparing this with the given expression \( h = \frac{nm^2g}{2Q^2B^2} \), we find that \( n = 4 \).

### Conclusion:
Thus, the value of \( n \) is \( 4 \).