Question

A particle with a mass of \(1.00 \mathrm{mg}\) and a charge \(q\) is moving at a speed of \(1000 . \mathrm{m} / \mathrm{s}\) along a horizontal path \(10.0 \mathrm{~cm}\) below and parallel to a straight current-carrying wire. Determine \(q\) if the magnitude of the current in the wire is \(10.0 \mathrm{~A}\).

Short answer

Answer: The charge of the particle is approximately 4.91 × 10^(-7) C.

Step-by-step solution

Find the gravitational force acting on the particle

First, let's calculate the gravitational force acting on the particle. The gravitational force can be found using the formula: $$F_g = mg$$ where \(F_g\) is the gravitational force, \(m\) is the mass of the particle, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{m/s^2}\)). Here, the mass of the particle is \(1.00 \mathrm{mg}\), which is equal to \(1.00 \times 10^{-3} \mathrm{kg}\). Therefore, the gravitational force acting on the particle is: $$F_g = (1.00 \times 10^{-3}\mathrm{kg})(9.81 \mathrm{m/s^2}) = 9.81 \times 10^{-3} \mathrm{N}$$

Find the magnetic field due to the current in the wire

In this step, we will find the magnetic field at the location of the particle due to the current in the wire. The magnetic field due to a straight current-carrying wire can be determined using Ampere's law: $$B = \frac{\mu_0I}{2 \pi r}$$ where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \mathrm{T\cdot m/A}\)), \(I\) is the current in the wire, and \(r\) is the distance between the wire and the point where the field is being measured. The current in the wire is \(10.0 \mathrm{A}\), and the distance between the particle and the wire is \(10.0 \mathrm{cm}\), which is equal to \(0.100 \mathrm{m}\). So, the magnetic field at the location of the particle is: $$B = \frac{4\pi \times 10^{-7} \mathrm{T\cdot m/A} \cdot 10.0 \mathrm{A}}{2 \pi \cdot 0.100 \mathrm{m}} = 2.00 \times 10^{-5} \mathrm{T}$$

Calculate the magnetic force acting on the particle

Now that we have the magnetic field at the location of the particle, we can calculate the magnetic force acting on the particle. As mentioned earlier, since the particle is not moving vertically, the magnetic force must be equal to the gravitational force acting on the particle. The magnetic force can be found using the formula: $$F = qvB$$ where \(F\) is the magnetic force, \(q\) is the charge, \(v\) is the velocity, and \(B\) is the magnetic field. We have the magnetic field, velocity, and the gravitational force which is equal to the magnetic force. So, we can rearrange the formula to find the charge of the particle: $$q = \frac{F}{vB}$$

Determine the charge of the particle

Now, we can plug in the values into the equation we derived in the previous step and find the charge of the particle: $$q = \frac{9.81 \times 10^{-3} \mathrm{N}}{(1000. \mathrm{m/s})(2.00 \times 10^{-5} \mathrm{T})} = 4.91 \times 10^{-7} \mathrm{C}$$ Thus, the charge of the particle is approximately \(4.91 \times 10^{-7} \mathrm{C}\).