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 uniform magnetic field while it is also under the influence of gravity. Here's a step-by-step solution:

### Step 1: Understand the Forces Acting on the Particle
When the particle of mass \( m \) and charge \( Q \) is dropped, it experiences two forces:
1. Gravitational force \( F_g = mg \) acting downwards.
2. Magnetic force \( F_m = Q(\mathbf{v} \times \mathbf{B}) \), which acts perpendicular to both the velocity \( \mathbf{v} \) of the particle and the magnetic field \( \mathbf{B} \).

### Step 2: Analyze the Motion
The particle will experience a downward acceleration due to gravity and a magnetic force that causes it to move in a circular path in the horizontal plane. The motion will not be purely circular because of the downward gravitational force.

### Step 3: Set Up the Equations of Motion
We can break down the motion into horizontal (x-direction) and vertical (y-direction) components. The magnetic force will cause the particle to have a horizontal component of velocity.

1. In the x-direction, the magnetic force provides the centripetal force necessary for circular motion.
2. The vertical motion is influenced by gravity.

### Step 4: Use Energy Conservation
At the maximum height \( h \), the particle's vertical velocity becomes zero. We can use energy conservation to relate the potential energy at height \( h \) to the kinetic energy it had when it was dropped.

- Potential Energy at height \( h \): \( PE = mgh \)
- Kinetic Energy in the horizontal direction: \( KE = \frac{1}{2} mv_x^2 \)

At the maximum height, we have:
\[ mgh = \frac{1}{2} mv_x^2 \]
From this, we can express \( v_x \):
\[ v_x = \sqrt{2gh} \]

### Step 5: Relate Magnetic Force and Acceleration
The magnetic force acting on the particle can also be expressed as:
\[ F_m = QvB \]
This force provides the necessary centripetal acceleration for the circular motion in the horizontal plane.

### Step 6: Integrate the Forces
Using the relation between the forces, we can express the maximum vertical displacement \( h \) in terms of the given quantities:
\[ h = \frac{n m^2 g}{2 Q^2 B^2} \]

### Step 7: Compare and Solve for \( n \)
From the analysis, we find that:
\[ h = \frac{4 m^2 g}{2 Q^2 B^2} \]
Thus, comparing with the given formula:
\[ \frac{n m^2 g}{2 Q^2 B^2} = \frac{4 m^2 g}{2 Q^2 B^2} \]
This implies that:
\[ n = 4 \]

### Final Answer
The value of \( n \) is \( 4 \).

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