2019 A/L Physics Past Paper · Question 48 (English Medium)

Answer: (4) Graph pair (4)

What the question asks

We want:

How the magnetic flux density $B$ changes between the two wires.

We do it for (a) and for (b).

Into the paper is positive.

Then we choose the right pair of graphs.

Relevant theory

At the distance $r$ from a long straight wire the flux density is:

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

The direction comes from the right hand screw rule. The field lines are circles round the wire.

The fields of two wires add as vectors.

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.

Applying it to this question

Two wires make the field.

The wire at $x = 0$ is at the distance $x$.

The wire at $x = d$ is at the distance $d - x$.

We write $k = \dfrac{\mu_0}{2\pi}$.

The sizes alone are:

$$B_1 = \dfrac{kI}{x}$$

$$B_2 = \dfrac{kI}{d - x}$$

Each one grows very large near its own wire.

Reasoning and calculation

The directions, from the right hand screw rule:

The wire at $x = 0$ has the current up. To its right the field goes into the paper, so it is $+$.

The wire at $x = d$ in (a) has the current up. To its left the field comes out of the paper, so it is $-$.

The wire at $x = d$ in (b) has the current down. To its left the field goes into the paper, so it is $+$.

(a) The same direction

$B_a(d/2) = 0$

The two fields point opposite ways, so we subtract.

$$B = kI\left(\dfrac{1}{x} - \dfrac{1}{d - x}\right)$$

Near $x = 0$ it is large and positive.

At $x = \dfrac{d}{2}$ it is zero.

Near $x = d$ it is large and negative.

The curve falls and changes sign.

(b) Opposite directions

$B_b(d/2) = 4kI/d$

The two fields point the same way, so we add.

$$B = kI\left(\dfrac{1}{x} + \dfrac{1}{d - x}\right)$$

Near both ends it is large and positive.

In the middle it is lowest, $\dfrac{4kI}{d}$, and it is above zero.

The curve is a U that does not touch the axis.

Reading the options

We need, for (a), a curve that goes from positive to negative and crosses zero in the middle.

We need, for (b), a U with a positive lowest point.

That is graph pair (4).

Why the other options fail

The two graphs are swapped.

In (b) the lowest point is shown as zero.

In (b) the sign is shown to change.

Checks

Symmetry check:

In (a), changing $x \to d - x$ gives $B \to -B$.

In (b), it gives $B \to B$.

The curves of graph pair (4) do both.

Conclusion

In (a) $B$ goes from positive to negative and is zero at $\dfrac{d}{2}$.

In (b) $B$ is a U shape above zero.

So the answer is option (4).