# Magnitude of magnetic field at different points near a capacitor.

• zenterix
zenterix
Homework Statement
Consider a capacitor that is charging, as in the picture below. The capacitor is ideal (no edge effects).

Points ##a## and ##b## areat a distance ##r_1>R## with respect to the center line, and ##c## and ##d## are at a distance ##r_2<R##.
Relevant Equations
Which of the following statements about ##B##, the magnitude of the magnetic field, at points ##a,b,c,## and ##d## are true?

##B(a)>B(b)##
##B(a)<B(b)##
##B(a)=B(b)##

##B(c)>B(d)##
##B(c)<B(d)##
##B(c)=B(d)##
Here is a picture depicting the capacitor and the points of interest.

I approached this problem by applying the Ampere-Maxwell law.

For each point I used an circular Amperian loop that I denote by ##P##, enclosing a circular surface ##S##.

Thus, for point ##b## we have

$$\oint_{P_b}\vec{B}\cdot d\vec{s}=B_b2\pi r_1=\mu_0 I\implies B_b=\frac{\mu_0 I}{2\pi r_1}$$

Similarly, for point ##d##

$$B_d=\frac{\mu_0 I}{2\pi r_2}$$

Next, I considered points ##a## and ##c##.

The magnitude of the electric field between the plates is ##\frac{q}{\epsilon_0 A}## where ##A## is the area of a capacitor plate.

$$\oint_{P_a}\vec{B}\cdot d\vec{s}=B_a2\pi r_1=\mu_0\frac{d}{dt}\left (\frac{q}{\epsilon_0 A}\right )\pi R^2$$

$$=\frac{\mu_0 I\pi R^2}{\epsilon_0\pi R^2}$$

$$\implies B_a=\frac{\mu_0I}{\epsilon_0 2\pi r_1}$$

Similarly

$$B_c2\pi r_2=\mu_0\frac{d}{dt}\left (\frac{q}{\epsilon_0 A}\right )\pi r_2^2=\frac{\mu_0 I\pi r_2^2}{\epsilon_0\pi R^2}$$

$$=\frac{\mu_0 Ir_2^2}{\epsilon_0 R^2}$$

$$B_c=\frac{\mu_0 I r_2}{\epsilon_0 2\pi R^2}$$

These are my calculations currently. They seem incorrect. Having a ##\epsilon_0## factor in the denominator of these expressions seems incorrect at first glance given that the order of magnitude of this constant is ##10^{-12}##.

Last edited:
zenterix said:
The magnitude of the electric field between the plates is ##\frac{q}{\epsilon_0 A}## where ##A## is the area of a capacitor plate.

$$\oint_{P_a}\vec{B}\cdot d\vec{s}=B_a2\pi r_1=\mu_0\frac{d}{dt}\left (\frac{q}{\epsilon_0 A}\right )\pi R^2$$
What is the fundamental law that you are using to set up this equation? You might be missing a factor of ##\epsilon_0##.

zenterix
If you are going to quote the Ampere-Maxwell law, quote it correctly. The displacement current term (Maxwell's correction) is ##\mathbf J_d=\dfrac{\partial \mathbf D}{\partial t}=\dfrac{\partial (\epsilon_0 \mathbf E)}{\partial t}##. The ##\epsilon_0## in the numerator cancels the one in the denominator.

zenterix
TSny said:
You might be missing a factor of ϵ0.
Indeed you are correct. I forgot a factor of ##\epsilon_0##.

$$\oint \vec{B}\cdot d\vec{s}=\mu_0\left (\iint_S \vec{J}\cdot\hat{n}dA+\epsilon_0\iint_S \vec{E}\hat{n}dA\right )$$

zenterix said:
Indeed you are correct. I forgot a factor of ##\epsilon_0##.

$$\oint \vec{B}\cdot d\vec{s}=\mu_0\left (\iint_S \vec{J}\cdot\hat{n}dA+\epsilon_0\iint_S \vec{E}\hat{n}dA\right )$$
OK, but here you forgot a time derivative.

zenterix
Indeed

$$\oint \vec{B}\cdot d\vec{s}=\mu_0\left (\iint_S \vec{J}\cdot\hat{n}dA+\epsilon_0\frac{d}{dt}\iint_S \vec{E}\hat{n}dA\right )$$

Fixing the original equations we have

$$B_a=B_b=\frac{\mu_0I}{2\pi r_1}$$

$$B_c=\frac{\mu_0I}{2\pi r_2}\frac{r_2^2}{R^2}$$

$$B_d=\frac{\mu_0I}{2\pi r_2}$$

Thus, ##B_a=B_b## and ##B_c<B_d##.

TSny
zenterix said:
Indeed
$$\oint \vec{B}\cdot d\vec{s}=\mu_0\left (\iint_S \vec{J}\cdot\hat{n}dA+\epsilon_0\frac{d}{dt}\iint_S \vec{E}\hat{n}dA\right )$$
Fixing the original equations we have

$$B_a=B_b=\frac{\mu_0I}{2\pi r_1}$$
$$B_c=\frac{\mu_0I}{2\pi r_2}\frac{r_2^2}{R^2}$$
$$B_d=\frac{\mu_0I}{2\pi r_2}$$
Thus, ##B_a=B_b## and ##B_c<B_d##.
Looks good to me.

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