2022 A/L Physics Past Paper · Question 50 (English Medium)

Answer: (2) Graph (2)

What the question asks

What we know:

A very long straight wire passes through the origin $O$, along the $Z$-axis. The current comes out of the paper, towards us.

The semicircle has radius $R$ and its centre is the wire. It is the upper half, on the side where $y$ is positive.

There are no numbers, so we keep the letters.

We want: the graph of $B_x$ against $x$ along the semicircle.

What each letter means:

$I$ ($\text{A}$): the current in the wire.

$R$ ($\text{m}$): the radius of the semicircle, the distance of every point of it from the wire.

$B_0$ ($\text{T}$): the size of the magnetic flux density at distance $R$ from the wire.

$B_x$ ($\text{T}$): the $x$-part of the magnetic flux density.

$x$, $y$ ($\text{m}$): the position of a point on the semicircle, measured from the wire.

Relevant theory

What is happening:

The wire makes a magnetic field that goes round it in circles.

Every point of the semicircle is the same distance $R$ from the wire. So the size of the field is the same, $B_0$, at every point of the semicircle.

The field points along the circle. With the current out of the paper, it goes round anticlockwise.

At the top of the semicircle, going anticlockwise means pointing to the left. At the right end it points up. At the left end it points down.

We want only the $x$-part of the field. So we look at how far the field points to the right or to the left.

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.

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

The first quote says that the field goes round the wire in circles.

The second quote gives the direction. A right handed corkscrew moving out of the paper turns anticlockwise when we look at it from the front. So the field goes anticlockwise.

The third quote gives the size, $\dfrac{\mu_0 I}{2\pi r}$. It depends on the distance $r$ from the wire and on nothing else about the point. On the semicircle $r = R$ everywhere.

For example, a roundabout in a playground that turns anticlockwise. A child at the top of it moves to the left. A child at its right edge moves straight up. Each child moves at the same speed, but the direction keeps turning.

Required equations

$B_0 = \dfrac{\mu_0 I}{2\pi R}$

$B_x = -B_0\,\dfrac{y}{R}$

$B_x \leq 0$

$B_x = -B_0 \text{ at } x = 0, \quad B_x = 0 \text{ at } x = \pm R$

Reasoning and calculation

$B_0$ is in $\text{T}$. The fraction $\dfrac{y}{R}$ has no unit, so $B_x$ is in $\text{T}$.

The minus sign comes from the direction. Anticlockwise at a point above the $x$ axis points a little to the left, towards $-x$.

$0$

$B_x = -B_0 \times \dfrac{0}{R}$

$-0.5 B_0$

$B_x = -B_0 \times 0.5$

$-B_0$

$B_x = -B_0 \times 1$

So $B_x$ goes from $0$ at the end, to below zero, to its lowest value $-B_0$ at the top, and back to $0$ at the other end. It never goes above zero.

That is a hollow below the axis, touching zero at $x = -R$ and at $x = R$. It is graph (2).

Check the five graphs.

Graph (1) stays constant, but the direction of the field keeps turning.

Graph (3) is above zero. That is for a current into the paper.

Graphs (4) and (5) change sign at $x = 0$. $B_x$ stays below zero on the whole semicircle.

Graph (2) is below zero, lowest at $x = 0$ and zero at both ends. It matches.

Checks

$-b0 * y / r$

Does it make sense?

The size of the whole field is the same all along the semicircle. The $x$-part is the whole field only at the top, where the field points left. At the two ends the field points straight up or straight down, and the $x$-part is zero. That fits graph (2).

The graph must be the same on the left and on the right of $x = 0$, because the field has the same size and the same tilt at two points the same distance from the $y$ axis. Graph (2) is.

Conclusion

$\text{graph }(2)$

$B_x$ is below zero, lowest at $x = 0$ and zero at $x = \pm R$. The graph is (2).

The answer is (2).

Find the direction first. Current out of the paper makes the field go anticlockwise round the wire.

The size is the same at the same distance. For the $x$-part, ask where the field points to the left, to the right, or straight up and down.

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