Question

g Three particles of equal charge, X, Y, and Z, enter a uniform magnetic field B. X has velocity of magnitude v parallel to the field. Y has velocity of magnitude v perpendicular to the field. Z has equal velocity components v parallel and perpendicular to the field. Rank the radii of their orbits from least to greatest.

Answer 1

Answer:

$R_Y<R_Z<R_X$ or radius of  Y< rad of Z< radius of X.

Explanation:

For a particle moving in a circular orbit in a magnetic field, the centripetal force must equal the magnetic force:

$\dfrac{mv^2}{R} = F_B$

where the magnetic force $F_B$ has the magnitude

$F_B =qvBsin(\theta)$;

therefore,

$\dfrac{mv^2}{R} = qvBsin(\theta)$

$R= \dfrac{mv}{qBsin(\theta)}.$

Now, for the particle X, $\theta =0$ since it is parallel to the magnetic field; therefore, it will experience no force and $R = \infty$.

For the particle, Y $\theta = 90^o$, and therefore, it will experience the greatest force, and thus its orbit will be the tightest (the radius will be the smallest)

Finally, for the particle Z, $0<\theta<90^o$ because its velocity will have both parallel and perpendicular components; therefore, the radius of its orbit will be greater than for particle Y, but less than for particle X.

Thus the radii, when ordered from least from greatest are

$R_Y<R_Z<R_X$.