Question

A long straight wire carries a \(4.70-\) A current in the positive \(x\) -direction. At a particular instant, an electron moving at $1.00 \times 10^{7} \mathrm{m} / \mathrm{s}\( in the positive \)y\( -direction is \)0.120 \mathrm{m}$ from the wire. Determine the magnetic force on the electron at this instant. Sce the figure with Problem \(65 .\)

Short answer

Question: An electron is moving in the positive y-direction at a speed of \(1.00 \times 10^7 \, \mathrm{m/s}\). A straight wire carrying a current of \(4.70 \, \mathrm{A}\) in the positive x-direction is located at a distance of \(0.120 \, \mathrm{m}\) away from the electron. Find the magnetic force acting on the electron. Answer: Follow the steps provided in the solution to calculate the magnetic force on the electron. The force vector has only a z-component, \(F_z = -qB_zv_y\), and its magnitude can be found by plugging in the known values as follows: \(F_z = -(-1.6 \times 10^{-19}\, \mathrm{C})\left(\dfrac{4\pi \times 10^{-7}\, \mathrm{T\cdot m/A}\cdot 4.70\, \mathrm{A}}{2\pi \cdot 0.120\, \mathrm{m}}\right)(1.00 \times 10^{7}\, \mathrm{m/s})\). Calculate the numerical value of the product to determine the force on the electron.

Step-by-step solution

Find the Magnetic Field

To find the magnetic field \(B\) due to a long straight wire with a known current, we'll use the Ampere's Circuital Law, which is given by \(B = \dfrac{\mu_0 I}{2\pi r}\), where \(\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\cdot m/A}\) is the permeability of free space, \(I\) is the current, and \(r\) is the distance from the wire. Use the given values for the current \(I = 4.70 \, \mathrm{A}\) and the distance \(r = 0.120 \, \mathrm{m}\) to calculate the magnetic field.

Calculate the Force on the Electron

Now we have the magnetic field \(B\), and we can calculate the force on the electron. First, let's write down the known values: \(q = -1.6 \times 10^{-19} \, \mathrm{C}\) is the charge of an electron and \(v = (0, 1.00 \times 10^{7}, 0) \, \mathrm{m/s}\) is its velocity vector. The magnetic field has only a z-component, i.e., \(B = (0, 0, B_z)\). We'll now use the formula for magnetic force \(F = q(v \times B)\), where \(\times\) is the vector cross product. Computing this cross product, we get \(F = (0, 0, -qB_zv_y)\), so the force has only a z-component.

Calculate the Magnitude of the Force

Now that we have found the force vector \(F = (0, 0, -qB_zv_y)\), let's calculate its magnitude. The only nonzero component of the force is its z-component, which is \(-qB_zv_y\). Plug in the known values for \(q\), \(B_z\), and \(v_y\) to get the magnitude of the force \(F_z = -qB_zv_y = -(-1.6 \times 10^{-19}\, \mathrm{C})\left(\dfrac{4\pi \times 10^{-7}\, \mathrm{T\cdot m/A}\cdot 4.70\, \mathrm{A}}{2\pi \cdot 0.120\, \mathrm{m}}\right)(1.00 \times 10^{7}\, \mathrm{m/s})\). Calculate the numerical value of the product and determine the force on the electron.