Question

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Answer given is 0.6 uN

Answer 1

By symmetry, I thought that the net force is 0. But the textbook says otherwise..

Answer 2

That's correct.
The vertical component of the field yields symmetric forces that cancel out.
But the radial ones add up to give an upward force of \(2\pi a.i.B\sin \theta\)

Answer 3

that does match with the answer


http://www.wolframalpha.com/input/?i=2*pi*1.8%2F100*4.6%2F1000*3.4%2F1000*sin%2820%29

Answer 4

But why \(\mu \times B\) is giving a different answer ?

thank you :)

Answer 5

Oh that is for torque experienced by the loop, gotcha !

Answer 6

Then, i guess, the ring experiences both torque and a net force upward

Answer 7

Yes, that's correct!
Torque is zero since B is symmetrical about the axis. There is only the upward net force.

Answer 8

I see... but the ring is in MAX potential energy state right ?

so if my interpretation is correct, if I slightly tilt the ring, the ring then experiences a torque and rotates by 180 degrees ?

Answer 9

\(B\) and \(\mu\) are almost 180 degrees out of phase in the given diagram..

Answer 10

here is my reasoning:
I write the differential force as below:
\[d{\mathbf{F}} = \frac{i}{c}d{\mathbf{s}} \times {\mathbf{B}}\]
now, I decompose the field as below:
\[{\mathbf{B}} = {B_z}{\mathbf{z}} + {B_r}{\mathbf{r}}\]
so, after a substitution, I get:
\[\begin{gathered}
d{\mathbf{F}} = \frac{i}{c}d{\mathbf{s}} \times \left( {{B_z}{\mathbf{z}} + {B_r}{\mathbf{r}}} \right) = \hfill \\
\hfill \\
= \frac{i}{c}d{\mathbf{s}} \times {B_z}{\mathbf{z}} + \frac{i}{c}d{\mathbf{s}} \times {B_r}{\mathbf{r}} \hfill \\
\end{gathered} \]

Answer 11

and the torque here acts to align \(B\) and \(\mu\) I guess

Answer 12

Correct, it means torque is zero, but any slight rotation will create a non zero torque that will increase and flip the ring upside down.
It is an unstable state of rotational equilibrium.

Answer 13

Thanks! I guess I am finally getting these :)

Answer 14

now, the first term has a contribute equal to zero, after integration, whereas the contribute of the scond term is:
\[F = \frac{{2\pi ia}}{c}B\sin 20\]
after integration

Answer 15

I have used the CGS system of units of measures

Answer 16

namely, \(c\) is the speed of light

Answer 17

Oh, I see... it is equivalent to \(dF = i dL\times B\) right ?

Answer 18

yes!

Answer 19

then you're integrating over the circumference of the ring, nice !

Answer 20

Yes, that's the starting point.

Answer 21

yes! Since I have integrate on the differential element \[d{\mathbf{s}}\]

Answer 22

integrated*

Answer 23

the resultant force is vertically directed, upward or downward depending on the orientation of the current

Answer 24

Don't you just apply F = BIL and then take the appropriate component sin(20)?
So it is B x I x 2 x pi x r x sin(20) = 6.05 x 10^(-7) N