Calculating Magnetic Field & Torque for a Circular Wire Loop

[lcam2]

Homework Statement

A circular wire loop of radius 19 cm carries a current of 16 A. A smaller flat coil of radius 0.76 cm, having 50 turns and a current of 1.2 A is concentric with the loop. The coil and loop are perpendicular.

a) What is the magnitude of the magnetic field that the loop alone produces at its center?

b) What is the magnitude of the torque that acts on the coil? (Assume the magnetic field due to the loop is essentially uniform throughout the volume occupied by the coil.)

Homework Equations

Biot-Savart Law
B= ($$\mu$$*I)/(2*Pi*r)

The Attempt at a Solution

I used biot-savart law, using I=16A, r=.19m, and $$\mu$$ =4pi *10^(-7)

I got 1.68e-5 but it is not the correct answer. I have no clue how to approach this problem,
Thanks in advance for any help

 

[lcam2]

A)
B=(4*10^-7)*(16)*Pi/(2(.19))

B=5.29*10^-5

B)
Torque = U x B
U= NIA
Torque = NIAB sin(90)
Torque = (50)(1.2)(pi)(.0076)^2(5.29E-5)= 5.57E-7

 

[Encephalon]

Your formula for the magnetic field at the center of a loop seems off. Consider the formula for a differential magnetic field from a differential length ds:

$$dB = \frac{ \mu_0}{4\pi }$$ $$\frac{ids\times \widehat{r}} {r^{2}}$$

Where $$\mu_0$$ is the magnetic constant. Remember that $$ids\times \widehat{r}} = ids|\widehat{r}|$$ because the circle's radius will always be perpendicular to it's length element.

Using that, you should get the right formula for the current at the center of a loop, which should give you the answer you need!

 

[lcam2]

Thanks a lot, i was missing a Pi.

 

[rwisz]

!



Great job using the Biot-Savart law! However, there are a few things to consider in order to arrive at the correct answer. First, it is important to note that the circular wire loop and the smaller flat coil are perpendicular to each other. This means that the magnetic field produced by the circular wire loop at the center will not contribute to the torque on the smaller coil. Therefore, you need to calculate the magnetic field produced by the smaller coil at its own center.

To do this, you can use the same Biot-Savart law equation, but with the new values for the smaller coil: I=1.2A, r=0.76cm, and \mu =4pi *10^(-7). This should give you a magnetic field of 1.27e-5 T at the center of the smaller coil.

Next, to calculate the torque on the smaller coil, you need to use the equation:

\tau = NIAcos(\theta)

Where N is the number of turns in the coil, I is the current, A is the area of the coil, and \theta is the angle between the magnetic field and the plane of the coil. In this case, N=50, I=1.2A, and A=\pi*r^2. Since the coil is perpendicular to the magnetic field, \theta = 90 degrees. Plugging in all the values, you should get a torque of 2.28e-5 Nm.

I hope this helps you arrive at the correct answers. Keep up the good work in your studies of magnetic fields and torque!