Question

As shown in the figure, a straight conductor parallel to the \(x\) -axis can slide without friction on top of two horizontal conducting rails that are parallel to the \(y\) -axis and a distance of \(L=0.200 \mathrm{~m}\) apart, in a vertical magnetic field of \(1.00 \mathrm{~T}\). A 20.0 - A current is maintained through the conductor. If a string is connected exactly at the center of the conductor and passes over a frictionless pulley, what mass \(m\) suspended from the string allows the conductor to be at rest?

Short answer

Answer: The mass that needs to be suspended from the string to keep the conductor at rest is approximately 0.408 kg.

Step-by-step solution

Identify the forces in the system

There are three forces acting on the system: 1. Magnetic force (\(F_{m}\)) on the conductor due to the current and magnetic field interaction 2. Gravitational force (\(F_{g}\)) on the mass due to gravity 3. Tension force (\(T\)) in the string Since the conductor is at rest, the sum of these forces must be equal to zero, i.e., \(F_{m} - T = 0\) and \(F_{g} - T = 0\).

Calculate magnetic force on the conductor

We can calculate the magnetic force on the conductor using the equation: \(F_{m} = B I L\) where \(B = 1.00 \ \text{T}\) is the magnetic field, \(I = 20.0 \ \text{A}\) is the current, and \(L = 0.200 \ \text{m}\) is the distance between the conducting rails. \(F_{m} = (1.00 \ \text{T})(20.0 \ \text{A})(0.200 \ \text{m}) = 4.00 \ \text{N}\)

Calculate the gravitational force acting on the mass

We can calculate the gravitational force acting on the mass using the equation: \(F_g = m g\) where \(m\) is the mass we need to find, and \(g = 9.81\ \text{m/s}^2\) is the acceleration due to gravity.

Calculate the tension in the string

As the conductor is at rest and the gravitational force acting on the mass is equal to the tension force, we have: \(T = F_g\) And since the magnetic force acting on the conductor is also equal to the tension force, we have: \(F_m = T\)

Solve for the mass

Now, substitute the magnetic force into the equation for the tension force: \(4.00\ \text{N} = m(9.81\ \text{m/s}^2)\) Solve for the mass: \(m = \frac{4.00\ \text{N}}{9.81\ \text{m/s}^2} = 0.408\ \text{kg}\) The mass that needs to be suspended from the string to keep the conductor at rest is approximately \(0.408\ \text{kg}\).