2019 A/L Physics Past Paper · Question 44 (English Medium)

The initial torque on wire loops: the order

Each of the single turn wire loops placed on $xy$ plane as shown in figures, carries the same current $I$ . A uniform magnetic field is applied along the positive direction of the $x$ -axis. Assume that each wire loop can rotate freely about its symmetric axis perpendicular to the magnetic field. Which choice represents the order of loops that the initial torque acting on them are in descending order?

(1) $P, Q, R, S$
(2) $R, Q, P, S$
(3) $Q, P, R, S$
(4) $S, R, Q, P$
(5) $R, Q, S, P$

Answer and explanation

Answer: (2) $R, Q, P, S$

What the question asks

We want:

The order of the four loops by the torque at the start.

The biggest torque comes first.

Relevant theory

The torque of a field $B$ on a loop of $N$ turns carrying a current $I$ is:

$\tau = NBIA\sin\alpha$

$\alpha$ is the angle between the field and the normal to the loop.

If the field lies in the plane of the loop, $\alpha = 90^\circ$ . The torque is the biggest, $BIA$ .

This is true for a loop of any shape. Only the area matters.

The couple on the coil causes it to rotate about the axis of suspension.

Required equations

$\tau = BIA$

Reading the options

The options are different orders of the four loops.

We find the areas and compare them.

Reasoning and calculation

What is the same:

The same $B$ .

The same $I$ .

One turn.

Every loop starts in the $xy$ plane, with the field in the plane.

So the torque depends on $A$ only.

What is different:

The shape and the size.

So only the area $A$ is different.

$P$ , the triangle

$A_P = 3$

The base goes from $x = 1$ to $3$ , so it is $2$ .

The height goes from $y = 1$ to $4$ , so it is $3$ .

$A = \tfrac{1}{2} \times 2 \times 3 = 3$ .

$Q$ , the rectangle

$A_Q = 4$

The width is $1$ , from $x = 1$ to $2$ .

The height is $4$ , from $y = 1$ to $5$ .

$A = 1 \times 4 = 4$ .

$R$ and $S$

$A_R \approx 7.07,\ A_S = 2.25$

$R$ is a circle. The diameter is $3$ , from $x = 1$ to $4$ . So the radius is $1.5$ .

$A_R = \pi \times 1.5^2 \approx 7.07$ .

$S$ is a square with side $1.5$ .

$A_S = 1.5^2 = 2.25$ .

Compare in pairs.

$7.07 > 4$ , so $R$ comes before $Q$ .

$4 > 3$ , so $Q$ comes before $P$ .

$3 > 2.25$ , so $P$ comes before $S$ .

The torque is proportional to the area.

$R\ (7.07) > Q\ (4) > P\ (3) > S\ (2.25)$ .

That is option (2).

Checks

$7.07$

$\tau_R = 0.5 \times 2 \times 2.25\pi$

$2.25$

$\tau_S = 0.5 \times 2 \times 2.25$

Conclusion

The first torque is $BIA$ .

It depends only on the area.

The order of the areas is $R, Q, P, S$ .

So the answer is option (2).

Thinking that the length or the perimeter of the loop decides the torque.

The torque is $BIA$ .

Only the area counts.

The shape does not matter.

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