There exists a long conductor along z-axis carrying a current of `I_(0)` along positive z-direction. A loop having total resistance R is placed symmetrical about x- and y-axes in xy plane as shown in figure. potential difference `v_(BA)=V` is applied. Radii of arcs are ‘a’ and ‘b’, respectively, as shown in the figure. The magnitude of force experienced by the arc MN is :

Options

A. zero
B. `(mu_(0)VI_(0))/(pi Rb)`
C. `(mu_(0)VI_(0))/(2 pi Rb)`
D. None of these

Explanation

To solve the problem of finding the magnitude of the force experienced by the arc MN in the given setup, we can follow these steps:

### Step 1: Understand the Setup
We have a long straight conductor along the z-axis carrying a current \( I_0 \) in the positive z-direction. There is also a loop in the xy-plane that is symmetrical about the x- and y-axes. The potential difference \( V_{BA} = V \) is applied across the loop, which has arcs of radii \( a \) and \( b \).

**Hint:** Visualize the setup and identify the direction of the current and the magnetic field produced by the straight conductor.

### Step 2: Determine the Magnetic Field
Using the right-hand rule, we can determine the direction of the magnetic field \( \mathbf{B} \) produced by the straight conductor. The magnetic field around a long straight conductor carrying current is given by:

\[ B = \frac{\mu_0 I_0}{2\pi r} \]

where \( r \) is the distance from the conductor to the point where we want to calculate the magnetic field.

**Hint:** Remember that the magnetic field circles around the conductor and its direction depends on the direction of the current.

### Step 3: Analyze the Current in the Arc MN
The potential difference \( V_{BA} \) applied across the loop causes a current \( I \) to flow through the loop. The direction of this current can be determined based on the potential difference. If \( V_A \) is at a higher potential than \( V_B \), the current will flow from A to B.

**Hint:** Use Ohm's law to relate the potential difference, current, and resistance: \( I = \frac{V}{R} \).

### Step 4: Calculate the Force on Arc MN
The force \( \mathbf{F} \) experienced by a current-carrying conductor in a magnetic field is given by:

\[ \mathbf{F} = I \mathbf{L} \times \mathbf{B} \]

where \( \mathbf{L} \) is the length vector of the current element and \( \mathbf{B} \) is the magnetic field vector. The magnitude of the force can be expressed as:

\[ F = B \cdot I \cdot L \cdot \sin(\theta) \]

In this case, since the magnetic field \( \mathbf{B} \) is perpendicular to the current \( I \) flowing through the arc MN, we need to consider the angle \( \theta \).

**Hint:** If the current direction is opposite to the magnetic field direction, \( \theta \) could be \( 180^\circ \), leading to \( \sin(180^\circ) = 0 \).

### Step 5: Conclusion
Since \( \sin(180^\circ) = 0 \), the force experienced by the arc MN is:

\[ F = B \cdot I \cdot L \cdot 0 = 0 \]

Thus, the magnitude of the force experienced by the arc MN is **0 Newtons**.

**Final Answer:** The correct option is **0**.