An infinte wire place along z-axis has current `I_(1)` in positive z-direction A conducting rod placed in xy plane parallel to y-axis has current `I_(2)` in positive y-direction The ends of the rod subtend `+30^(@)` and `-60^(@)` at the origin with positive x direction The rod is at a distance a from the origin. Find net force on the rod .

Options

A. `F=(mu_(0))/(4pi)l_(1)l_(2)ln 3(-hatk)`
B. `F=(mu_(0))/(4pi)2l_(1)l(2)ln3(-hati)`
C. `F=(mu_(0))/(4pi)2l_(1)l_(2)ln3(hatk)`
D. `F=(mu_(0))/(4pi)2l_(1)l_(2)ln3(hatk)`

Answer

To solve the problem of finding the net force on the conducting rod due to the infinite wire, we will follow these steps:

### Step 1: Understand the Geometry
The infinite wire is placed along the z-axis and carries a current \( I_1 \) in the positive z-direction. The conducting rod is in the xy-plane, parallel to the y-axis, carrying a current \( I_2 \) in the positive y-direction. The ends of the rod subtend angles of \( +30^\circ \) and \( -60^\circ \) at the origin with respect to the positive x-direction.

### Step 2: Determine the Length of the Rod
Let the distance from the origin to the rod be \( a \). The rod extends from the angle \( -60^\circ \) to \( +30^\circ \). The length of the rod can be determined using the angles:
- The coordinates of the endpoints of the rod can be found using trigonometry:
- For the endpoint at \( +30^\circ \):
\[ L_1 = a \cos(30^\circ) = a \frac{\sqrt{3}}{2} \]
- For the endpoint at \( -60^\circ \):
\[ L_2 = a \cos(-60^\circ) = a \frac{1}{2} \]
- The length of the rod \( L \) is the difference between these two lengths:
\[ L = L_1 - L_2 = a \frac{\sqrt{3}}{2} - a \frac{1}{2} = a \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) \]

### Step 3: Calculate the Force on the Rod
The force on a current-carrying conductor in a magnetic field is given by:
\[ F = I_2 L B \]
Where \( B \) is the magnetic field due to the infinite wire. The magnetic field at a distance \( r \) from a long straight wire carrying current \( I_1 \) is given by:
\[ B = \frac{\mu_0 I_1}{2 \pi r} \]
In our case, \( r = a \).

### Step 4: Substitute and Integrate
The force on the rod can be expressed as:
\[ F = I_2 L \left(\frac{\mu_0 I_1}{2 \pi a}\right) \]
Now, substituting the expression for \( L \):
\[ F = I_2 \left(a \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)\right) \left(\frac{\mu_0 I_1}{2 \pi a}\right) \]
The \( a \) cancels out:
\[ F = I_2 \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) \left(\frac{\mu_0 I_1}{2 \pi}\right) \]

### Step 5: Final Expression
Now, simplifying the expression:
\[ F = \frac{\mu_0 I_1 I_2}{2 \pi} \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) \]
Calculating \( \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3} - 1}{2} \):
\[ F = \frac{\mu_0 I_1 I_2}{2 \pi} \cdot \frac{\sqrt{3} - 1}{2} \]

### Step 6: Conclusion
Thus, the net force on the rod is:
\[ F = \frac{\mu_0 I_1 I_2 (\sqrt{3} - 1)}{4 \pi} \]