EMI BYU 9

9. Equation of Particle Path

Step 1: Initial Impulse (Betatron Effect)

As the magnetic field $B$ ramps up from 0 to $B_0$, an induced electric field $E$ accelerates the particle. Using Faraday’s law for the circular loop at radius $r=a$:

$$ \oint E \cdot dl = -\frac{d\Phi}{dt} \implies E(2\pi a) = -\pi a^2 \frac{dB}{dt} \implies E = -\frac{a}{2} \frac{dB}{dt} $$

The impulse imparted is $F \Delta t = q E \Delta t$:

$$ m v_0 = q \int E dt = q \left( \frac{a}{2} \right) B_0 \implies v_0 = \frac{q a B_0}{2m} $$

The direction is tangential (along -y axis for the given setup).

Step 2: Motion in Constant Field

After the ramp-up, the field is constant $B_0$. The particle moves with velocity $v_0$ perpendicular to $B_0$, resulting in circular motion.

Radius of curvature $R_c$:

$$ R_c = \frac{m v_0}{q B_0} = \frac{m}{q B_0} \left( \frac{q a B_0}{2m} \right) = \frac{a}{2} $$

Step 3: Equation of Path

The particle starts at $(a, 0)$ and moves vertically. Since the radius is $a/2$, the circle must be tangent to the vertical line $x=a$ at $(a,0)$. The center is at $(a/2, 0)$.

Equation of a circle with center $(h, k) = (a/2, 0)$ and radius $r = a/2$:

$$ (x – h)^2 + (y – k)^2 = r^2 $$