2020 A/L Physics Past Paper · Question 12 (English Medium)

Answer: (4) $40\ \mu\text{T}\ \downarrow$

What the question asks

We must find: the magnitude and direction of the magnetic flux density produced at the centre $C$ by the currents in the eight wires.

Relevant theory

Then the iron powder will be found to set around the wire along concentric circular paths indicating the presence of a magnetic field and also its structure.

When a right handed corkscrew is passed in the direction of an electric current passing through a conductor, the sense of rotation of the screw represents the direction of the magnetic field formed around the conductor.

Magnetic flux density at a point around a very long straight conductor carrying a current.

Applying it to this question

The flux density at distance $r$ from a long straight wire:

$$B = \left(\frac{\mu_0}{4\pi}\right)\frac{2I}{r}$$

Contributing elements: eight wires. The two middle wires ($\otimes$ left, $\odot$ right) and the top/bottom middle wires ($\otimes$, $\otimes$) are each $10\ \text{cm}$ from $C$; the four corner wires are at $10\sqrt2\ \text{cm}$.

Direction notation: in the plane of the page, "up" ($\uparrow$), "down" ($\downarrow$). By the corkscrew rule, the field circles clockwise around a current going into the page ($\otimes$) and anticlockwise around one coming out ($\odot$); each wire's field is perpendicular to the line from the wire to $C$.

Reasoning and calculation

Diagonal pairs, and the top–bottom pair

$B = 0$

The top-left $\otimes$ and bottom-right $\otimes$: equal distance from $C$, opposite sides, same current direction — their fields at $C$ are equal in magnitude, opposite in direction. The top-right $\odot$/bottom-left $\odot$ $pair\,and\,the\,top/bottom\,middle$ $\otimes$ pair cancel the same way.

Middle left wire ($\otimes$)

$B_1 = 20\ \mu\text{T}\ \downarrow$

$C$ is $10\ \text{cm}$ to its right; the clockwise field there points downward: $B_1 = \left(\dfrac{\mu_0}{4\pi}\right)\dfrac{2I}{r} = 10^{-7} \times \dfrac{2 \times 10}{0.10}$.

Middle right wire ($\odot$)

$B_2 = 20\ \mu\text{T}\ \downarrow$

$C$ is $10\ \text{cm}$ to its left; the anticlockwise field there also points downward; same magnitude.

$2 \times 10^{-5}$

$B_1 = 10^{-7} \times \frac{2 \times 10}{0.10}$

$40$

$B = B_1 + B_2 = 20 + 20$

Resultant: apart from the three cancelling pairs, only the two middle wires remain: at $C$, $B = 40\ \mu\text{T}$, downward ($\downarrow$).

Reading the options

Comparing with the options: $40\ \mu\text{T}\ \downarrow$ — option (4).

Why the other options fail

Counts only one middle wire.

Misses both the direction and the magnitude.

Reverses the sense of the corkscrew rule.

Takes the field along the wire–$C$ line.

Checks

$2 * 1e-7 * 2 * I / r$

Symmetry check: rotating $180°$ about $C$ maps the grid onto itself — wire pairs with the same current direction always cancel at $C$: three such pairs here. The remaining two have opposite current directions, so their fields add; a second way, summing all eight components separately, also gives $B_x = 0$, $B_y = -40\ \mu\text{T}$ ✓.

Conclusion

$B = 40\ \mu\text{T}\ \downarrow$

With three pairs cancelling, the two middle wires each give $20\ \mu\text{T}$ downward at $C$: $B = 40\ \mu\text{T}\ \downarrow$ — the answer is (4).

Drawing each wire's field pointing from the wire toward the point.

A straight wire's field lines are circles around it: the field is perpendicular to the wire–point line.