2020 A/L Physics Past Paper · Question 44 (English Medium)

Answer: (2) $B_1 \uparrow$, $B_2 \odot$

What the question asks

We must find: the directions of $B_1$, $B_2$ for the electron to go straight in region I and turn upward on a semicircle in region II.

Relevant theory

The electric field intensity at a certain point in an electric field is the force acting ona unit positive (+) charge placed at that point.

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.

Fleming's left-hand rule: with the fore finger, middle finger and thumb of the left hand held mutually perpendicular — fore finger along the field, middle finger along the current — the thumb gives the direction of the force.

the charge carriers of the current, the electrons would drift in the same path in the direction opposite to the conventional direction of the current J.

Applying it to this question

Direction notation: in the plane of the page, right, left, up ($\uparrow$), down ($\downarrow$); perpendicular to the page, out ($\odot$), in ($\otimes$). The electron moves to the right entering both regions.

Forces: in region I, the electric force $F_E = eE$ and the magnetic force $F_B = evB_1$; in region II, only the magnetic force $evB_2$. The electron carries charge $-e$: an electron moving right is equivalent to a conventional current flowing left.

Reasoning and calculation

Direction rule: since $E = \dfrac{F}{q}$ with $q = -e$, the electric force on the electron is opposite to $E$ — out of the page ($\odot$). For a straight path, $evB_1$ must be into the page ($\otimes$) to balance $eE$.

Left-hand rule: middle finger (current) left, thumb (force) into the page, so the fore finger (field) points up $\Rightarrow B_1 \uparrow$.

Region II: the centre of the semicircle is above the entry point, so the force on entry must point up; middle finger left, thumb up, so the fore finger points out of the page $\Rightarrow B_2 \odot$.

$evB_1 = eE$

$v = \frac{E}{B_1}$

$evB_2 = \frac{mv^2}{r}$

Final directions: $B_1$ in the plane, pointing up ($\uparrow$); $B_2$ out of the page ($\odot$).

Reading the options

Comparing with the table: $B_1 \uparrow$, $B_2 \odot$ — option (2).

Why the other options fail

$B_2 \otimes$: the electron would curve downward.

$B_1 \odot$: the magnetic force lies in the plane and cannot cancel the electric force.

$B_1 \otimes$: the force lies in the plane.

$B_1 \downarrow$: the magnetic force would add to the electric force instead of cancelling it.

Checks

Second way (vector product): $\vec F = q\,\vec v \times \vec B$, $q = -e$, $\vec v = v\,\hat x$. For $\vec B_1 = B_1\hat y$: $\hat x \times \hat y = \hat z$ (out of the page), and since $q < 0$, $\vec F$ points into the page ✓ — opposite to the electric force.

For $\vec B_2 = B_2\hat z$: $\hat x \times \hat z = -\hat y$, and since $q < 0$, $\vec F$ points up ($+\hat y$) ✓.

Conclusion

$B_1$ in the plane, pointing up ($\uparrow$): $evB_1$ into the page cancels $eE$; $B_2$ out of the page ($\odot$): it turns the electron upward — the answer is (2).

Using the electron's direction of motion as the current direction in the left-hand rule.

The electron carries negative charge: the (conventional) current is opposite to its motion — or find the force first and reverse its direction.