Question

A thin wire oriented along the x-axis carries a current I in the +x direction. At a certain instant, a point charge +q is coincident with the point (x, y, z)= (0, a, 0) and moves with velocity v= v 3 [2ˆx+2ˆy−zˆ]. What is the vector force F on the charge at that instant, in terms of I, q, v, a, and fundamental constants?

Answer 1

Answer:

The force on the charge is caused by the magnetic field created by the current in the wire. So, first we need to calculate the magnitude and direction of the magnetic field, then we will use this magnetic field to calculate the force on the charge.

Ampere's Law will help us to find the magnetic field created by the wire.

$\mu_0 I_{enc} = \int {B} \, dr\\ \mu_0 I = B2\pi a\\B = \frac{\mu_0 I}{2\pi a}$

The direction of the B-field can be found by right-hand rule. Point your thumb in the direction of the current. Curl your fingers around the wire. The direction that the four fingers point is the direction of the B-field, it is around the wire.

At the point (0,a,0) the B-field is directed towards +z-direction.

So,

$\vec{B} = \frac{\mu_0 I}{2\pi a}\^{z}$

Now, we can calculate the force by using the following formula:

$\vec{F} = q\vec{v} \times \vec{B}$

In order to do the cross product, we should use determinant.

$\vec{F} = q \left[\begin{array}{ccc}\^{x}&\^{y}&\^{z}\\v_x&v_y&v_z\\B_x&B_y&B_z\end{array}\right] = q \left[\begin{array}{ccc}\^{x}&\^y&\^z\\2v^3&2v^3&-v^3\\0&0&\frac{\mu_0 I}{2\pi a}\end{array}\right] \\\\= \^{x}(2v^3 \frac{\mu_0 I}{2\pi a}q) - \^{y}(2v^3 \frac{\mu_0 I}{2\pi a}q)\\\vec{F} = \frac{2v^3q\mu_0 I }{2\pi a}[\^x + \^y]$