A rectangular coil with 20 windings carries a current of 2.00 mA flowing in the counterclockwise direction. It has two sides that are parallel to the \(y\) -axis and have length \(8.00 \mathrm{~cm}\) and two sides that are parallel to the \(x\) -axis and have length \(6.00 \mathrm{~cm} .\) A uniform magnetic field of \(50.0 \mu \mathrm{T}\) acts in the positive \(x\) -direction. What torque must be applied to the loop to hold it steady?

Given: - Number of windings: \(n = 20\) - Current: \(I = 2.00 \,\text{mA} = 2.00\times10^{-3}\,\text{A}\) - Length of sides parallel to the \(y\)-axis: \(l_y = 8.00 \,\text{cm} = 0.08 \,\text{m}\) - Length of sides parallel to the \(x\)-axis: \(l_x = 6.00 \,\text{cm} = 0.06 \,\text{m}\) - Uniform magnetic field strength: \(B = 50.0\,\mu\text{T} = 50.0\times10^{-6}\,\text{T}\)

Since the magnetic field is parallel to the \(x\)-axis, there will be no force exerted on the sides parallel to the \(y\)-axis (the force will be perpendicular to length and field). Therefore, we can move on to the next step.

The magnetic force on a current-carrying wire can be calculated using the formula: $$F = nIBL\sin\theta$$ As the magnetic field is parallel to the \(x\)-axis, it will be perpendicular to the sides of the loop parallel to the \(x\)-axis, making the angle \(\theta = 90^\circ\). Therefore, the magnetic force on these sides is: $$F_{x} = nIBl_y\sin(90^\circ) = 20 \times 2.00\times10^{-3} \text{A} \times 50.0\times10^{-6} \text{T} \times 0.08 \text{m}$$ $$F_{x} = 1.6\times10^{-5} \,\text{N}$$

To find the net torque on the loop, which must be counteracted to hold the loop steady, we need to calculate the torque on both sides parallel to the \(x\)-axis and add them up. The torque is given by the cross product: $$\tau = r \times F_{x}$$ Since \(F_{x}\) acts equally on both sides, we can calculate the torque on one side and simply double it. The torque on one side is: $$\tau_{one-side} = \frac{l_x}{2} \times F_{x}$$ $$\tau_{one-side} = 0.5 \times 0.06\,\text{m} \times 1.6\times10^{-5}\,\text{N} = 4.8\times10^{-7} \,\text{N}\cdot\text{m}$$ Now, double the torque to account for both sides: $$\tau_{total} = 2 \times \tau_{one-side} = 2 \times 4.8\times10^{-7}\,\text{N}\cdot\text{m} = 9.6\times10^{-7} \,\text{N}\cdot\text{m}$$

The torque that must be applied to hold the rectangular coil steady in the given uniform magnetic field is: $$\tau_{total} = 9.6\times10^{-7}\,\text{N}\cdot\text{m}$$