2020 A/L Physics Past Paper · Question 37 (English Medium)

Answer: (2) $\dfrac{r}{\sqrt2}$

What the question asks

We must find: the radius $r'$ of the particle's new circular path in the same magnetic field, after it loses half its kinetic energy passing through the foil.

Relevant theory

No force acts on an electric charge which is at rest in a magnetic field. However if the charge is moving it is equivalent to an electric current and hence a force would act on it.

Since a body is moving with an acceleration only under the action of an external force, a body performing circular motion is doing so under the action of an external force acting in the direction of the centripetal acceleration. This force is thus known as "centripetal force" and is given by.

$$F = \frac{mv^2}{r} = mr\omega^2$$

$$W = \frac{1}{2}mv^2$$

Required equations

$F = Bqv$

$Bqv = \frac{mv^2}{r}$

$r = \frac{mv}{qB}$

$E_k = \frac{1}{2}mv^2$

Reasoning and calculation

What changes, what stays the same: $m$, $q$, $B$ are the same before and after the foil.

Only the speed $v$ changes (the foil takes away energy).

Since the magnetic force is always perpendicular to the velocity, the field itself never changes the energy.

Proportion: in $r = \dfrac{mv}{qB}$, $\dfrac{m}{qB}$ is constant, so $r \propto v$.

From $E_k = \dfrac12 mv^2$, $v \propto \sqrt{E_k}$.

Combining the two, $r \propto \sqrt{E_k}$.

$\frac{r'}{r} = \frac{v'}{v} = \sqrt{\frac{E_k'}{E_k}}$

$E_k' = \frac{1}{2}E_k$

$0.707$

$\frac{r'}{r} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt2}$

$r' = \frac{r}{\sqrt2}$

Reading the options

Comparing with the choices: $\dfrac{r}{\sqrt2}$ — option (2).

Checks

$sqrt(k)$

Dimension check:

the unit of $\dfrac{mv}{qB}$ is $\dfrac{\text{kg m s}^{-1}}{\text{C} \cdot \text{T}} = \dfrac{\text{kg m s}^{-1}}{\text{C} \cdot \text{N A}^{-1}\text{m}^{-1}} = \text{m}$ ✓ (since $\text{C} = \text{A s}$).

Second way (momentum): $r = \dfrac{p}{qB}$, $p = \sqrt{2mE_k}$.

Halving $E_k$ makes $p$ $\dfrac{1}{\sqrt2}$ times, so $r$ too ✓.

Conclusion

$r' = \frac{r}{\sqrt2}$

$r \propto v \propto \sqrt{E_k}$; half the energy gives $r' = \dfrac{r}{\sqrt2}$ — the answer is (2).

Taking the speed as halved when the kinetic energy is halved.

$E_k \propto v^2$: the speed falls to only $\dfrac{1}{\sqrt2}$ times.