D.3 Motion in electromagnetic fields

The magnetic force on a moving charge is given by $F=qvB\sin \theta$ where $\theta$ is the angle between the velocity vector and the magnetic field direction.

Here $\theta =30°$, $q=e$ (the elementary charge), so the force is $F=evB\sin 30°$ This is neither the maximum force (which requires $\theta =90°$) nor zero.

Since the velocity is not perpendicular or parallel to the field, the $\sin \theta$ factor must be included. The correct expression is $F=evB\sin 30°$.

We need the magnetic force when the particle moves at $30°$ to the field. The general formula is $F=qvB\sin \theta$.

The option $F=evB$ corresponds to $\sin \theta =1$, i.e. $\theta =90°$. Since the angle here is $30°$, not $90°$, this is incorrect.

The formula uses $\sin \theta$, not $\cos \theta$. Using cosine would apply to work calculations or component projections, not to the magnetic force magnitude.

The force is zero only when the velocity is parallel (or anti-parallel) to the field, i.e. $\theta =0°$ or $180°$. Here $\theta =30°$, so the force is non-zero.

The remaining option $F=evB\sin 30°$ correctly applies the formula $F=qvB\sin \theta$ with $q=e$ and $\theta =30°$.

When a charged particle moves perpendicular to a uniform magnetic field, it follows a circular path. Setting the magnetic force equal to the centripetal force gives $R=\frac{mv}{qB}$

$R=\frac{(1.67\times 10^{-27})(3.0\times 10^6)}{(1.6\times 10^{-19})(0.40)}$

Numerator: $1.67\times 10^{-27}\times 3.0\times 10^6=5.01\times 10^{-21}$ N·s. Denominator: $1.6\times 10^{-19}\times 0.40=6.4\times 10^{-20}$ C·T.

$R=\frac{5.01\times 10^{-21}}{6.4\times 10^{-20}}\approx 0.078\text{ m}$

Use $R=\frac{mv}{qB}$ and compute the numerical result to match one of the options.

This would result from using $B=0.5$ T or making an arithmetic error in the denominator. With $B=0.40$ T, the denominator is $6.4\times 10^{-20}$, which does not yield $0.031$ m.

This is exactly double the correct answer, suggesting an error such as forgetting a factor in the denominator (e.g. using $q/2$) or doubling the numerator.

This could arise from using $v=2.0\times 10^6$ m s${}^{-1}$ instead of $3.0\times 10^6$ m s${}^{-1}$, which is not the value given.

The calculation $\frac{5.01\times 10^{-21}}{6.4\times 10^{-20}}\approx 0.078$ m matches the first option.