2021 A/L Physics Past Paper · Question 33 (English Medium)

Answer: (1) $a$

What the question asks

We want: the direction of the total magnetic field at the corner $D$.

Three long wires stand at three corners of a square. Each wire carries the same current $I$. The current comes out of the paper, towards you.

The plan: first find what each wire does at $D$ on its own. Then add the three.

Relevant theory

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.

According to the structure of the composite field, the two fields have added with each other to form a stronger field on the left side of the conductor while on the right side of the conductor the two fields oppose each other resulting a weaker field.

Applying it to this question

Think of a roundabout in a park. Children run round and round it.

Each wire is like a roundabout. Its magnetic field goes round and round the wire, in circles.

At any spot, the field points along the circle. It never points straight at the wire or straight away from it.

A long wire makes this field at a distance $r$:

$$B = \dfrac{\mu_0 I}{2\pi r}$$

In words: a spot that is farther from the wire gets a weaker field.

Which way does each field point at $D$? Use the corkscrew rule. The current comes out of the paper, so the field goes round each wire anticlockwise.

Now look at the three wires one by one.

Wire $A$

$B_1$, down and to the right

Wire $A$ is a distance $x$ from $D$, up and to the right. Its field is $B_A = \dfrac{\mu_0 I}{2\pi x}$. Call this $B_1$.

Going anticlockwise round $A$, the field at $D$ points down and to the right.

Wire $C$

$B_1$, down and to the left

Wire $C$ is also a distance $x$ from $D$, down and to the right. Its field is $B_C = B_1$.

Going anticlockwise round $C$, the field at $D$ points down and to the left.

Wire $B$

$0.71\,B_1$, straight down

Wire $B$ is across the square, a distance $\sqrt{2}\,x$ from $D$, straight to the right. It is farther away, so it is weaker: $B_B = \dfrac{B_1}{\sqrt{2}} = 0.71\,B_1$.

Going anticlockwise round $B$, the field at $D$ points straight down.

Reasoning and calculation

Now add the three. Split each push into a sideways part and a down part.

Sideways: $A$ pushes right and $C$ pushes left, both by $0.71\,B_1$. It is like two children pulling a rope with the same strength in opposite ways. Nothing moves. They cancel.

Down: wire $A$ gives $0.71\,B_1$, wire $C$ gives $0.71\,B_1$ and wire $B$ gives $0.71\,B_1$. All three point down, so they add.

$0$

$0.71\,B_1 - 0.71\,B_1$

$2.12\,B_1$

$0.71\,B_1 + 0.71\,B_1 + 0.71\,B_1$

The sideways part is zero. The down part is $2.12\,B_1$.

So the total field at $D$ points straight down. Straight down is arrow $a$.

Reading the options

The arrows are: $a$ down, $b$ down and right, $c$ right, $d$ up and right, $e$ up.

Straight down is $a$. This is option (1).

Why the other options fail

Arrow $b$ is the push of wire $A$ only. Wire $C$ cancels its sideways part.

The field of a wire goes round the wire. It never points along the line to the wire.

Arrow $d$ points straight at wire $A$. A wire's field goes round it.

Arrow $e$ is the opposite way. The field goes anticlockwise round a wire with the current coming out of the paper.

Checks

Does it make sense?

Look at the picture. Wires $A$ and $C$ are the same distance from $D$, one above and one below.

So their sideways pushes must be equal and opposite. Only a push straight down (or straight up) can be left.

$3 * a / b$

Conclusion

$B_D = 3 \times 0.71\,B_1 = 2.12\,B_1 \text{ (down)}$

The sideways parts of $A$ and $C$ cancel. The down parts of $A$, $B$ and $C$ add up to $2.12\,B_1$.

The total field at $D$ points straight down, along arrow $a$. The answer is (1).

Each wire's field goes round the wire. Use the corkscrew rule for the way.

Split every push into sideways and down parts, and add the parts one direction at a time.

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