Question

A stationary circular wall clock has a face with a radius of 15 cm. Six turns of wire are wound around its perimeter; the wire carries a current of 2.0 A in the clockwise direction. The clock is located where there is a constant, uniform external magnetic field of magnitude 70 mT (but the clock still keeps perfect time). At exactly 1:00 P.M., the hour hand of the clock points in the direction of the external magnetic field. (a) After how many minutes will the minute hand point in the direction of the torque on the winding due to the magnetic field? (b) Find the torque magnitude.

Short answer

1. The interval after which the minute hand will point in the direction of the torque on the winding due to the magnetic field is 20min.
2. The magnitude of the torque is$\tau =5.9\times {10}^{-2}N\cdot m$.

Step-by-step solution

Given

1. Current through the wire,i=2.0A.
2. The radius, r=15cm=0.15m
3. The number of turns,N=6.
4. Uniform external magnetic field,$B=70mT=70\times {10}^{-3}T$
5. At exactly 1:00 pm, the hour hand of the clock points in the direction of the external magnetic field.

Determine the formula for the torque and the magnetic moment

By using the vector cross product $\mu ⃗\times B⃗$and applying the right-hand rule, we can find theinterval after which the minute hand will point in the direction of the torque on the winding due to the magnetic field. By using the formula for the magnitude of the torque, we can find the magnitude of the torque.

Formula:

1. The torque is given by$\tau ⃗=\mu ⃗\times B⃗$
2. The magnitude of the torque is given by$\tau ⃗=\left|\mu ⃗\times B⃗\right|$
3. The magnitude of magnetic moment is$\mu =NiA$

(a) Calculate the interval after which the minute hand will point in the direction of the torque on the winding due to the magnetic field.

Consider the torque is given by:

$\tau ⃗=\mu ⃗\times B⃗$

Since the current goes clockwise around the clock, $\overrightarrow{\mu }$points to the wall.

Since role="math" localid="1662907647478" $\overrightarrow{B}$points toward one-hour or 5-minute mark, by the property of the vector product, $\overrightarrow{\tau }$must be perpendicular to it.

Thus, by using the right-hand rule, we can say that$\tau$points at the 20-minute mark.

So, the time interval is 20min.

(b) Calculate the magnitude of the torque.

The magnitude of the torque is given by

$$\begin{gathered} \tau =|\mu ⃗\times B⃗| \\ \tau =\mu Bsin{90}^0 \\ \tau =\mu B \end{gathered}$$

…… (1)

But, the magnitude of magnetic moment is

$\mu =NiA$

With$A=\pi r^2$,

$\mu =\pi Ni{r}^2$

Substitute the values in equation

$\tau =\pi Ni{r}^{2}B$

Substitute the values and solve as:

$\tau =\pi \times 6\times (2.0\hspace{0.33em}A)\times (0.15\hspace{0.33em}m{)}^2\times (70\times {10}^{-3}T)$

$\tau =5.9\times {10}^{-2}N\cdot m$