Calculating Magnetic Field & Torque for a Circular Wire Loop

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 7K views
lcam2
Messages
28
Reaction score
0

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= ([tex]\mu[/tex]*I)/(2*Pi*r)


The Attempt at a Solution


I used biot-savart law, using I=16A, r=.19m, and [tex]\mu[/tex] =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
 
Physics news on Phys.org
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
 
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:

[tex]dB = \frac{ \mu_0}{4\pi }[/tex] [tex]\frac{ids\times \widehat{r}} {r^{2}}[/tex]

Where [tex]\mu_0[/tex] is the magnetic constant. Remember that [tex]ids\times \widehat{r}} = ids|\widehat{r}|[/tex] 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!
 
Thanks a lot, i was missing a Pi.