GSO001 Problem 13

The magnetic flux density $B$ at distance $r$ from a long straight current $I$ :

B = \frac{\mu_0 I}{2\pi r}

Wire 1 at $x = d$ : Distance from observation point $(x, 0, z)$ is $d - x$ . By the right-hand rule, it creates a field in the $+y$ direction.

B_1 = \frac{\mu_0 I}{2\pi (d - x)}

Wire 2 at $x = -d$ : Distance is $d + x$ . It creates a field in the $-y$ direction.

B_2 = \frac{\mu_0 I}{2\pi (d + x)}

The $y$ -component of the combined field:

B_y = B_1 - B_2 = \frac{\mu_0 I}{2\pi} \frac{2x}{d^2 - x^2} = \frac{\mu_0 I x}{\pi (d^2 - x^2)}

Since $|x| \ll d$ , we approximate $d^2 - x^2 \approx d^2$ :

B_y \approx \frac{\mu_0 I}{\pi d^2} x

With particle velocity $\vec{v} = (v_x, 0, v_0)$ and $\vec{B} = (0, B_y, 0)$ :

\vec{v} \times \vec{B} = (-v_0 B_y, 0, v_x B_y)

The $x$ -component of the Lorentz force:

F_x = -q v_0 B_y = -q v_0 \left( \frac{\mu_0 I}{\pi d^2} x \right) = - \left( \frac{q v_0 \mu_0 I}{\pi d^2} \right) x

This is a restoring force of the form $F_x = -Kx$ , which causes simple harmonic motion in the $x$ -direction.

(The $z$ -force $F_z = q v_x B_y$ is a second-order small quantity, justifying the approximation that $v_z \approx v_0$ = constant.)

The equation of motion in the $x$ -direction:

m \frac{d^2 x}{dt^2} = - \left( \frac{q v_0 \mu_0 I}{\pi d^2} \right) x

Comparing with $m a = -m \omega^2 x$ , the angular frequency $\omega$ :

\omega = \sqrt{\frac{q v_0 \mu_0 I}{\pi m d^2}}

The period $T = 2\pi / \omega$ :

T = 2\pi d \sqrt{\frac{\pi m}{q v_0 \mu_0 I}}

L = v_0 T = 2\pi d \sqrt{\frac{\pi m v_0}{q \mu_0 I}}

Substituting numerical values. First, compute the content under the square root:

Numerator:

\pi m v_0 = \pi \times (4.0 \times 10^{-5}) \times (1.0 \times 10^2) = 4.0 \pi \times 10^{-3}

Denominator:

q \mu_0 I = (2.0 \times 10^{-4}) \times (4\pi \times 10^{-7}) \times 50 = 4.0\pi \times 10^{-9}

Content inside the square root:

\frac{4.0\pi \times 10^{-3}}{4.0\pi \times 10^{-9}} = 10^6

Therefore:

L = 2\pi \times 0.50 \times \sqrt{10^6} = 2\pi \times 0.50 \times 1000 = 1000\pi

Answer: 1000

Motion of a Charged Particle Between Two Parallel Currents