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? \( \bullet \)

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 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 Field Strength:** The magnetic field strength \( B \) is given as \( 0.4 \, \text{T} \). 4. **Force Calculation:** The force acting on the lower half of the loop is: \[ F = I \cdot L \cdot B = I \cdot (0.5) \cdot (0.4) \] \[ F = 0.2 I \, \text{N} \] 5. **Equilibrium Condition:** When the current flows through the loop, it creates a force that needs to be balanced by the weight added to the right arm of the scale. The weight of 25 g can be converted to Newtons: \[ \text{Weight} = m \cdot g = 0.025 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 0.24525 \, \text{N} \] Setting the magnetic force equal to the weight added: \[ 0.2 I = 0.24525 \] 6. **Solving for Current \( I \):** \[ I = \frac{0.24525}{0.2} = 1.22625 \, \text{A} \] Therefore, the current flowing in the loop is approximately: \[ I \approx 1.23 \, \text{A} \] ### Part (b) If the direction of the current is reversed, the direction of the magnetic force will also reverse (according to the right-hand rule). This means that instead of pushing the scale down on the left side, it will now push it up. To bring the scale back to equilibrium, you would need to **remove the 25 g weight** that was added to the right arm of the scale. This is because the force due to the reversed current will now act in the opposite direction, effectively reducing the weight on the left side of the scale. ### Summary: (a) The current flowing in the loop is approximately \( 1.23 \, \text{A} \). (b) If the current direction is reversed, you would need to remove the 25 g weight from the right arm to bring the scale back to equilibrium.