2018 A/L Physics Past Paper · Question 48 (English Medium)

Ten wires — magnetic force on the centre wire

An infinitely long thin straight wire held perpendicular to the plane of the paper at point $O$ carries a current $I$ into the paper as shown in the figure. Nine other similar infinitely long wires, parallel to the above wire, and held on the circumference of a circle of radius $r$ with centre at point $O$ , each carries a current $I$ into the paper. Except for wires $A$ and $B$ , the angular separation between any two consecutive wires is $30^\circ$ as shown. The magnitude and the direction of the magnetic force per unit length on the wire held at the centre $O$ due to other wires are, (Take $\cos 30^\circ = \frac{\sqrt3}{2}$ .)

(1) $\dfrac{\mu_0 I^2}{2\pi r}(1 + \sqrt3)$ in the direction of $YO$ .
(2) $\dfrac{\mu_0 I^2}{2\pi r}(1 + \sqrt3)$ in the direction of $OY$ .
(3) $\dfrac{\mu_0 I^2}{\pi r}(1 + \sqrt3)$ in the direction of $OY$ .
(4) $\dfrac{\mu_0 I^2}{2r}(1 + \sqrt3)$ in the direction of $OX$ .
(5) $\dfrac{3\mu_0 I^2}{2\pi r}$ in the direction of $YO$ .

Answer and explanation

Answer: (2) $\dfrac{\mu_0 I^2}{2\pi r}(1 + \sqrt3)$ in the direction of $OY$ .

What the question asks

We want: the resultant magnetic force per unit length on the wire at $O$ . Both its size and its direction.

Relevant theory

Two parallel conductors are carrying currents in the same direction. Consider the composite magnetic field around the two conductors.

This results in an attraction between the two conductors due to thrusts from outside.

Applying it to this question

Nine wires are at the same distance $r$ from $O$ .

Seven are on the upper half: $0^\circ$ , $30^\circ$ , $60^\circ$ , $90^\circ$ , $120^\circ$ , $150^\circ$ , $180^\circ$ .

Two are below: $A$ at $210^\circ$ and $B$ at $330^\circ$ .

$f = \frac{\mu_0 I \cdot I}{2\pi r} = \frac{\mu_0 I^2}{2\pi r}$

All the currents go into the paper. Currents in the same direction attract.

So each force points from $O$ towards that wire.

Reasoning and calculation

The left side is a mirror image of the right side.

So the sideways ( $x$ ) parts of the forces cancel in pairs.

Only the up and down ( $y$ ) parts are left.

The $0^\circ$ and $180^\circ$ wires

$0$

They pull along the $x$ axis in opposite directions.

The pulls are equal. So they cancel completely.

The $30^\circ$ and $150^\circ$ wires

$f(\tfrac12 + \tfrac12) = f$

Their sideways parts cancel.

Each one has an up part $f\sin30^\circ$ .

The $60^\circ$ and $120^\circ$ wires

$f\sqrt3$

Each one has an up part $f\sin60^\circ$ .

The $90^\circ$ wire

$f$

It is straight above $O$ .

Its whole force $f$ points up.

Wires $A$ and $B$

$-f$

Each one has a down part $f\sin30^\circ$ .

$f(0 + \sin30^\circ + \sin60^\circ + 1 + \sin60^\circ + \sin30^\circ + 0)$

$f\left(\tfrac12 + \tfrac{\sqrt3}{2} + 1 + \tfrac{\sqrt3}{2} + \tfrac12\right) = f(2 + \sqrt3)$

$f(\sin210^\circ + \sin330^\circ) = f\left(-\tfrac12 - \tfrac12\right) = -f$

$F = f(2 + \sqrt3 - 1) = \frac{\mu_0 I^2}{2\pi r}(1 + \sqrt3)$

$F = \frac{\mu_0 I^2}{2\pi r}(1 + \sqrt3)$

It points up, in the direction of $OY$ .

Reading the options

This size and this direction are option (2).

Why the other options fail

This option says the force is along $YO$ , downwards.

This option has $\pi r$ below the line instead of $2\pi r$ . That is twice the right size.

This option says the force is along $OX$ , and $\pi$ is missing.

This option says $3$ units along $YO$ .

Checks

$2 + sqrt(3) - 1$

Seven wires pull up and only two pull down.

So it makes sense that the centre wire is pulled up, along $OY$ .

Conclusion

$F = \frac{\mu_0 I^2}{2\pi r}(1 + \sqrt{3})$

$F = \frac{\mu_0 I^2}{2\pi r}(1 + \sqrt3)$

It is along $OY$ .

The answer is option (2).

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