A square frame carrying a current `I = 0.9 A` is located in the same plane as a long straght wire carrying a current, `I_(0) = 5.0 A`. The frame side has a length `a = 8.0 cm`. The axis of the frame passing thorugh the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is `eta = 1.5` times greater than the side of the frame. FInd:
(a) Ampere force acting on the frame,
(b) the mechnical work to be performed in order to turn the frame throguh `180^(@)` about its axis, with the currents maintained constant.

Answer

To solve the problem, we will break it down into two parts: (a) calculating the Ampere force acting on the square frame, and (b) determining the mechanical work required to turn the frame through 180 degrees about its axis while maintaining the currents constant.

### Given Data:
- Current in the square frame, \( I = 0.9 \, A \)
- Current in the straight wire, \( I_0 = 5.0 \, A \)
- Side length of the square frame, \( a = 8.0 \, cm = 0.08 \, m \)
- Distance between the wire and the frame, \( d = 1.5 \times a = 1.5 \times 0.08 \, m = 0.12 \, m \)

### Part (a): Ampere Force Acting on the Frame

1. **Calculate the Magnetic Field due to the Straight Wire:**
The magnetic field \( B \) at a distance \( d \) from a long straight wire carrying current \( I_0 \) is given by:
\[ B = \frac{\mu_0 I_0}{2\pi d} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A \) (permeability of free space).

Substituting the values:
\[ B = \frac{4\pi \times 10^{-7} \times 5.0}{2\pi \times 0.12} = \frac{2 \times 10^{-6}}{0.12} \approx 1.67 \times 10^{-5} \, T \]

2. **Calculate the Force on Each Side of the Frame:**
The force \( F \) on a current-carrying conductor in a magnetic field is given by:
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
where \( \theta \) is the angle between the current direction and the magnetic field. For the sides of the frame parallel to the wire, \( \theta = 90^\circ \) (so \( \sin(90^\circ) = 1 \)).

Each side of the frame has a length \( a \), so the force on the two sides parallel to the wire is:
\[ F_{\text{parallel}} = 2 \cdot I \cdot a \cdot B \]

Substituting the values:
\[ F_{\text{parallel}} = 2 \cdot 0.9 \cdot 0.08 \cdot 1.67 \times 10^{-5} \approx 2.4 \times 10^{-6} \, N \]

3. **Calculate the Net Force:**
The forces on the two sides of the frame that are perpendicular to the wire will cancel each other out since they are equal in magnitude and opposite in direction. Therefore, the net force acting on the frame is:
\[ F_{\text{net}} = F_{\text{parallel}} = 2.4 \times 10^{-6} \, N \]

### Part (b): Mechanical Work to Turn the Frame

1. **Calculate the Torque on the Frame:**
The torque \( \tau \) on a current loop in a magnetic field is given by:
\[ \tau = n \cdot I \cdot A \cdot B \]
where \( n \) is the number of turns (1 for a single loop), \( A \) is the area of the loop, and \( B \) is the magnetic field.

The area \( A \) of the square frame is:
\[ A = a^2 = (0.08)^2 = 0.0064 \, m^2 \]

Substituting the values:
\[ \tau = 1 \cdot 0.9 \cdot 0.0064 \cdot 1.67 \times 10^{-5} \approx 8.06 \times 10^{-10} \, N \cdot m \]

2. **Calculate the Work Done to Rotate the Frame:**
The work done \( W \) to rotate the frame through an angle \( \theta \) is given by:
\[ W = \tau \cdot \theta \]
where \( \theta \) must be in radians. For \( 180^\circ \):
\[ \theta = \pi \, \text{radians} \]

Substituting the values:
\[ W = 8.06 \times 10^{-10} \cdot \pi \approx 2.53 \times 10^{-9} \, J \]

### Final Answers:
(a) The Ampere force acting on the frame is approximately \( 2.4 \times 10^{-6} \, N \).

(b) The mechanical work to be performed in order to turn the frame through \( 180^\circ \) is approximately \( 2.53 \times 10^{-9} \, J \).