8. A square loop of side \( a=50 \mathrm{~cm} \) is suspended from the left arm of a scale, which is kept at equilibrium by adding weights to the right side. The lower half of the loop is in a uniform magnetic field of magnitude 0.4 T , orthogonal to the surface of the loop, but the upper half is outside the magnetic field. (a) It is observed that when a current \( I \) flows in the loop, one needs to add 25 g to the right arm to bring the scale back to equilibrium. Find the current flowing in the loop. (b) If the direction of current were now to be reversed, what would need to be done to bring the scale back to equilibrium? \( \cdot \)

To solve this problem, we will use the concept of the magnetic force acting on a current-carrying conductor in a magnetic field. ### Part (a) 1. **Magnetic Force on the Loop:** The magnetic force \( F \) on a segment of the loop that is in the magnetic field can be calculated using the formula: \[ F = I \cdot L \cdot B \cdot \sin(\theta) \] where: - \( I \) is the current flowing through the loop, - \( L \) is the length of the segment of the loop in the magnetic field, - \( B \) is the magnetic field strength, - \( \theta \) is the angle between the direction of the current and the magnetic field. In this case, the lower half of the square loop is in the magnetic field, and the current flows in a direction that is perpendicular to the magnetic field. Therefore, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \). 2. **Length of the Segment in the Magnetic Field:** The length \( L \) of the lower half of the square loop is equal to the side length \( a \): \[ L = a = 0.5 \, \text{m} \] 3. **Magnetic Force Calculation:** The magnetic force acting on the lower half of the loop is: \[ F = I \cdot L \cdot B = I \cdot (0.5 \, \text{m}) \cdot (0.4 \, \text{T}) = 0.2 I \, \text{N} \] 4. **Weight Added to the Scale:** The weight added to the right arm of the scale is 25 g, which can be converted to Newtons: \[ W = m \cdot g = 0.025 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 0.24525 \, \text{N} \] 5. **Setting Forces Equal:** To bring the scale back to equilibrium, the magnetic force must equal the weight added: \[ 0.2 I = 0.24525 \] 6. **Solving for Current \( I \):** \[ I = \frac{0.24525}{0.2} = 1.22625 \, \text{A} \] Thus, the current flowing in the loop is approximately: \[ \boxed{1.23 \, \text{A}} \] ### Part (b) If the direction of the current is reversed, the direction of the magnetic force will also reverse. This means that instead of pushing the loop upwards, the magnetic force will now push it downwards. To bring the scale back to equilibrium after reversing the current, we would need to **remove the 25 g weight** that was added to the right arm of the scale. This is because the downward magnetic force will counteract the weight of the loop, and thus, to maintain equilibrium, we need to remove the weight. In summary, to bring the scale back to equilibrium after reversing the current, we need to: \[ \text{Remove the 25 g weight.} \]