Current `I_(0)` flows through solenoid of length L having N number of turns, when it is connected to DC emf. If charged particle is projected along the axis of solenoid with a speed `v_(0)`, then the force on the charged particle in the solenoid:

Options

A. Becomes zero
B. Remains same
C. Decreases
D. Increases

Answer

To solve the problem step-by-step, we need to analyze the situation involving a charged particle moving through a solenoid with a current flowing through it.

### Step 1: Understand the Setup
We have a solenoid of length \( L \) with \( N \) turns, and it carries a current \( I_0 \). A charged particle is projected along the axis of the solenoid with an initial speed \( v_0 \).

### Step 2: Determine the Magnetic Field Inside the Solenoid
The magnetic field \( B \) inside a long solenoid is given by the formula:
\[ B = \mu_0 \frac{N}{L} I_0 \]
where \( \mu_0 \) is the permeability of free space.

### Step 3: Identify the Direction of the Magnetic Field
The direction of the magnetic field inside the solenoid is along the axis of the solenoid, following the right-hand rule.

### Step 4: Analyze the Motion of the Charged Particle
The charged particle is moving along the axis of the solenoid with speed \( v_0 \). Since both the velocity \( v_0 \) of the charged particle and the magnetic field \( B \) are in the same direction (along the axis of the solenoid), we can analyze the force acting on the charged particle.

### Step 5: Calculate the Magnetic Force on the Charged Particle
The magnetic force \( F \) on a charged particle moving in a magnetic field is given by the formula:
\[ F = q(v \times B) \]
where \( q \) is the charge of the particle, \( v \) is the velocity vector, and \( B \) is the magnetic field vector.

### Step 6: Determine the Angle Between \( v \) and \( B \)
In this case, since the charged particle is moving along the axis of the solenoid and the magnetic field is also along the same axis, the angle \( \theta \) between \( v \) and \( B \) is \( 0^\circ \).

### Step 7: Calculate the Sine of the Angle
The sine of the angle \( \theta \) is:
\[ \sin(0^\circ) = 0 \]

### Step 8: Conclude the Force Calculation
Substituting \( \sin(0^\circ) = 0 \) into the force equation:
\[ F = qvB \sin(0^\circ) = qvB \cdot 0 = 0 \]
Thus, the force acting on the charged particle inside the solenoid is zero.

### Final Answer
The force on the charged particle in the solenoid is:
\[ F = 0 \]

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