A charge particle moves in a circle around an infinite line charge with the center of circle at the line and line being perpendicular to the plane of circle. Let r be the radius of circle. The velocity of the particle depends upon which power of r.

Options

A. 1
B. 2
C. `-1`
D. none of these

Explanation

To solve the problem, we need to analyze the motion of a charged particle moving in a circle around an infinite line charge. Let's break down the solution step by step.

### Step 1: Understand the Setup
We have a charged particle moving in a circular path around an infinite line charge. The line charge is oriented perpendicular to the plane of the circle, and the radius of the circle is denoted as \( r \).

### Step 2: Determine the Electric Field
The electric field \( E \) due to an infinite line charge with linear charge density \( \lambda \) at a distance \( r \) from the line charge is given by the formula:
\[ E = \frac{2k\lambda}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) is Coulomb's constant.

### Step 3: Calculate the Force on the Charged Particle
The force \( F \) acting on a charged particle with charge \( q \) in the electric field \( E \) is given by:
\[ F = qE = q \left(\frac{2k\lambda}{r}\right) = \frac{2kq\lambda}{r} \]

### Step 4: Apply Centripetal Force Condition
For the charged particle to move in a circle, the centripetal force must equal the electric force acting on it. The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

Setting the electric force equal to the centripetal force, we have:
\[ \frac{2kq\lambda}{r} = \frac{mv^2}{r} \]

### Step 5: Simplify the Equation
We can cancel \( r \) from both sides (assuming \( r \neq 0 \)):
\[ 2kq\lambda = mv^2 \]

### Step 6: Solve for Velocity \( v \)
Rearranging the equation to solve for \( v \), we get:
\[ v^2 = \frac{2kq\lambda}{m} \]
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{2kq\lambda}{m}} \]

### Step 7: Analyze the Dependence on \( r \)
From the final expression for \( v \), we see that the velocity \( v \) does not depend on \( r \) at all. Therefore, we conclude that the velocity of the charged particle is independent of the radius \( r \).

### Final Answer
The velocity of the particle depends on \( r^0 \) (which means it does not depend on \( r \)).

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