Q58P - Fundamentals Of Physics

Question

The capacitor in Fig. 32-7 is being charged with a $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{A}$  current. The wire radius is $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{mm}$, and the plate radius is $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{cm}$ . Assume that the current i in the wire and the displacement current id in the capacitor gap are both uniformly distributed. What is the magnitude of the magnetic field due to i at the following radial distances from the wire’s center: (a) $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mm}$  (inside the wire), (b)  $\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mm}$ (outside the wire), and (c)  $\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{ }\mathbf{cm}$ (outside the wire)? What is the magnitude of the magnetic field due to id at the following radial distances from the central axis between the plates: (d)  $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mm}$ (inside the gap), (e) $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{mm}$  (inside the gap), and (f)   (outside the gap)? (g) $\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{ }\mathbf{cm}$ Explain why the fields at the two smaller radii are so different for the wire and the gap but the fields at the largest radius are not?

Step-by-Step Solution

Answer

1. Magnetic field at $1.00 \mathrm{mm}$  inside the wire is $222 \mathrm{\mu T}$
2. Magnetic field at  $3.00 \mathrm{mm}$ outside the wire is  $167\mathrm{\mu T}$
3. Magnetic field at $2.20 \mathrm{cm}$  outside the wire is  $22.7 \mathrm{\mu T}$
4. Magnetic field due to ${\mathrm{i}}_{\mathrm{d}}$  at $1.00 \mathrm{mm}$  inside the gap is  $1.25 \mu T$
5. Magnetic field due to $i_d$  at  $3.00$ mm inside the gap is  $3.75 \mathrm{\mu T}$
6. Magnetic field due to  $i_d$ at  $2.20 \mathrm{cm}$ outside the gap is  $22.7\mathrm{\mu T}$
7. Because displacement current in the gap is spread over a larger cross sectional area, values of B within that area are relatively small. Outside that cross section, the two values of B are identical.

Step 1: Listing the given quantities

Charged current is  $\mathrm{i}=2.50 \mathrm{A}$

Wire radius is $\mathrm{R}=1.50 \mathrm{mm}$ 

Plate radius is  $\mathrm{r}=2.00 \mathrm{cm}$

Step 2: Understanding the concepts of magnetic field

Here we need to use the equation for magnetic field due to charged wire and magnetic field due to charge between parallel plates.

Formula:

 

 $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{ir}}{2{\mathrm{\pi R}}^2}$

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}$

Step 3: (a) Calculations of the Magnetic field at 1 . 00   mm inside the wire

Using the formula given below,

 $\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{ir}}{2{\mathrm{\pi R}}^2}\\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 2.50\times 2.50\times {10}^{-2}}{2\mathrm{\pi }\times {(1.50\times {10}^{-3})}^2}\\ =2.22\times {10}^{-4} \mathrm{T}\\ =222 \mathrm{\mu T}\end{array}$

Magnetic field at  $1.00 \mathrm{mm}$ inside the wire is $222 \mathrm{\mu T}$

Step 4: (b) Calculations of the magnetic field at 3 . 0   mm outside the wire

Use the following formula:

 $\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}\\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 2.50}{2\mathrm{\pi }\times 3\times {10}^{-3}}\\ =1.667\times {10}^{-4} \mathrm{T}\\ =167 \mathrm{\mu T}\end{array}$

Magnetic field at  $3.00 \mathrm{mm}$ outside the wire is  $167\mathrm{\mu T}$

Step 5: (c) Calculations of the Magnetic field at 2 . 20 cm outside the wire

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}}\\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 2.50}{2\mathrm{\pi }\times 2.20\times {10}^{-2}}\\ =2.27\times {10}^{-5} \mathrm{T}\\ =22.7\times {10}^{-6} \mathrm{T}\\ =22.7 \mathrm{\mu T}\end{array}$

Magnetic field at $2.20 \mathrm{cm}$  outside the wire is  $22.7 \mathrm{\mu T}$

Step 6: (d) Calculations of the Magnetic field due to i d at 1 . 00   mm inside the gap

Use the following formula:

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{d}}\mathrm{r}}{2{\mathrm{\pi R}}^2}\\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 2.50\times 1\times {10}^{-3}}{2\mathrm{\pi }\times {(2\times {10}^{-2})}^2}\\ =1.25\times {10}^{-6} \mathrm{T}\\ =1.25 \mathrm{\mu T}\end{array}$

Magnetic field due to  $i_d$ at $1.00 \mathrm{mm}$  inside the gap is  $1.25 \mathrm{\mu T}$

Step 7: (e) Calculations of the magnetic field due to i d at 3 . 0 mm inside the gap

Use the following formula:

 $\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{ir}}{2{\mathrm{\pi R}}^2}\\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 2.50\times 3\times {10}^{-3}}{2\mathrm{\pi }\times {(2\times {10}^{-2})}^2}\\ =3.75\times {10}^{-6} \mathrm{T}\\ =3.75 \mathrm{\mu T}\end{array}$

Magnetic field due to $i_d$  at  $3.00$ mm inside the gap is  $3.75 \mathrm{\mu T}$

Step 8: (f) Calculations of the magnetic field due to i d at 2 . 20   cm outside the gap

Use the following formula:

 $\begin{array}{c}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{ir}}{2{\mathrm{\pi R}}^2}\\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times 2.50}{2\mathrm{\pi }\times (2.2\times {10}^{-2})}\\ =2.27\times {10}^{-5} \mathrm{T}\\ =22.7 \mathrm{\mu T}\end{array}$

Magnetic field due to ${\mathrm{i}}_{\mathrm{d}}$  at $2.20 cm$   outside the gap is  $22.7\mathrm{\mu T}$

Step 9: (g) Explanation

Why the field at two smaller radii is so different?

Because displacement current in the gap is spread over a larger cross-sectional area, values of B within that area are relatively small. Outside that cross section, the two values of B are identical.