# Magnitude of magnetic field at the center of circular wire

I'm preparing for an exam by solving the sample questions , here is the one I'm having difficulty with :

Following is the given circuit.
Which contains two resistance $R_1$ and $R_2$ in form of circle of radius $r = 1 m$ with a battery having e.m.f. $V = 10π$ volt.
Upper resistance is having resistivity $p_1 = 4$ & lower resistance having resistivity $p_2 = 2$.
Angle between two points $A$ and $B$ is $60°$. (wires have same cross section $A_1 = A_2 = 2 cm^2$)

Diagram : Here is the question based on this passage that I'm having difficulty with :
Find magnitude of magnetic field at center.

(A) 0T (B) 1T (C) 2T (D) πT

The problem is I don't know any formula to calculate the magnetic field at a point nor am I expected to know (probably , at least it is not in syllabus) .
Also what is 'T' ? Is it unit of magnetic field?
The answer is 0 so I guess I should be able to reason why it should be 0 but I don't know how. Please help.

• I've looked at this and I must admit I don't get the value zero. Are you sure the resistivity of the upper and lower wires is different? – John Rennie Sep 27 '13 at 14:49
• PS the symbol T is for Tesla, which is the SI unit of magnetic field strength - en.wikipedia.org/wiki/Tesla_(unit) – John Rennie Sep 27 '13 at 16:51

This is how (I think) the calculation is done, though I don't get the field strength to be zero. There is a nice description of this on the Hyperphysics site.

For any random piece of wire you get the field strength at a point using the Biot-Savart law. For a circular arc the field strength at the centre of the circle works out very easily because the wire is always normal to the line joining it to the centre (i.e. a radius) so the BS law simplifies to:

$$dB = \frac{\mu_0}{4\pi} \frac{I}{r^2} dl$$

where $I$ is the current, $r$ the radius of the circle and $dl$ the infinitesimal piece of the wire. Everything on the right hand side is constant except for $dl$, so integrating just gives you the length of the arc, $l$, so:

$$B = \frac{\mu_0}{4\pi} \frac{I}{r^2} l$$

The current $I$ is just $V/R$, and the resistance $R$ is given by:

$$R = \frac{\rho l}{A}$$

where $\rho$ is the resistivity and $A$ the area of the wire, so the current, $V/R$, is:

$$I = \frac{VA}{\rho l}$$

Substitute this into the expression above for $B$ and we get:

\begin{align} B &= \frac{\mu_0}{4\pi} \frac{1}{r^2} \frac{VA}{\rho l} l \\ &= \frac{\mu_0}{4\pi} \frac{VA}{r^2 \rho} \end{align}

So, interestingly, the length of the arc, $l$, doesn't appear in the expression for the magnetic field. This seems odd, but it's because two effects cancel out. If you make the length of the arc longer this would increase $B$, but the increased resistance reduces the current and this cancels the increase in length leaving $B$ unchanged.

Anyhow, in your circuit the upper and the lower arc produce fields in different directions, i.e. $B$ and $-B$. if the wires were identical then the lengths of the arcs wouldn't matter and the fields would always sum to zero. The problem is that the resistivity of the upper wire is twice that of the lower wire - $\rho_1 = 2\rho_2$. That means the total field in the centre will be:

\begin{align} B_{total} &= \frac{\mu_0}{4\pi} \frac{VA}{r^2 \rho_2} - \frac{\mu_0}{4\pi} \frac{VA}{r^2 2\rho_2} \\ &= \frac{\mu_0}{4\pi} \frac{VA}{r^2 2\rho_2} \end{align}

and this value is independant of the angle subtended by the upper arc. I get this to be $5\pi \times 10^{-11}$T or about $1.6 \times 10^{-10}$ Tesla.

Incidentally, one Tesla is a hugely strong field. Even the enormously powerful magnets in the LHC are only about 9T. It should be obvious that a 31V battery isn't going to produce a 1T field at a distance of a metre, so of the options given (A) is the only physically reasonable one. This makes me wonder if you have made a mistake copying the problem.

• I'm sure I haven't made a mistake in copying. I rechecked the question . But thanks for the answer. – A Googler Sep 28 '13 at 6:55

Well I'm a bit confused by the question itself....."Magnitude of magnetic field at the center of circular wire."

Is the question really what is the field at the center of a circular wire LOOP ; because that could be different from the field at the center of a circular wire.

And then that is a rather strange looking 60 degree angle; it looks more like 120 degrees to me.

In which case one arc is twice the length of the other arc so the shorter arc carries twice the current of the longer arc (it has half the resistance) and since the current flow directions are opposite, the two arc fields exactly cancel out at the center of the loop.

So the answer IS 0 Tesla, and the wire resistivity IS the same for both arcs.

And is this a "study question" or is it a homework (or hardware) question that I am not allowed to answer ?

But if that short arc angle really is 60 degrees, and not 120 degrees, then the whole problem is different again.

And on one final note; the battery Voltage is not specified, nor the resistivity of the wire, in which case, the currents flowing are completely unknown except for their ratio.

Ergo, the correct answer cannot be anything other than zero Tesla, because none of the other answers could be proven, given an entirely unknown current.

Well I see I was looking at the information in the diagram, but apparently the Voltage is known, and the resistivities ; so you can follow the simple method I used and their values.

It would be nice if the numbers and the diagram agreed; when they don't you can't believe either of them.

• Not homework. It is from a sample question paper for an exam . – A Googler Sep 28 '13 at 7:00

C(circumference) = 2PI*R = 2PI m R1 = p1*l/A => 4* (60/360)C / 2(cm)^2 = 4(1/6)*2PI / 0.0002(m^2) = 2122 ohm R2 = p2*l/A => 2* (300/360)C / 2(cm)^2 = 2(5/6)*2PI / 0.0002(m^2) = 5304 ohm

I1 = v/R1 = 10PI volt/ 2122 ohm = 0.0592 A I2 = v/R2 = 10PI volt/ 5304 ohm = 0.0236 A

B = U0*I/4PI *(Integral(dl/r^2)) ---- U0 => permeability of free space, I => current

B1 = 4*PI*10^-7 * 0.0592 A * 1/6 * 2PI / 1^2 = 19.72 * 10^-10 B2 = 4*PI*10^-7 * 0.0236 A * 5/6 * 2PI / 1^2 = 39.2 * 10^-10

B(total) = B2 - B1 = 2 nano Tesla OUT OF THE PAGE

for the magnetic field to be zero, the two wires must have the same Resistivity which is not so in this case.

• Welcome, Natan. We have the MathJax rendering engine running on the site, which allows you to write math in a LaTeX-like language and have it rendered neatly.For instance B_2 - B_1 = 2 \,\mathrm{nT} placed between dollar signs ($) renders as$B_2 - B_1 = 2 \,\mathrm{nT}\$. Using this tool would greatly improve the readability of your post. – dmckee --- ex-moderator kitten Sep 27 '13 at 18:39