Q58P The capacitor in Fig. 32-7 is be... [FREE SOLUTION]

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Chapter 32: Q58P (page 970)

The capacitor in Fig. 32-7 is being charged with a $2.50 A$ current. The wire radius is $1.50 mm$ , and the plate radius is $2.00 cm$ . 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鈥檚 center: (a) $1.0 mm$ (inside the wire), (b) $3.0 mm$ (outside the wire), and (c) role="math" localid="1662982620568" $2.20 cm$ (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) $1.0 mm$ (inside the gap), (e) $3.00 mm$ (inside the gap), and (f) (outside the gap)? (g) $2.20 cm$ 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?

Short Answer

Expert verified

Magnetic field at $1.00 mm$ inside the wire is $222 渭罢$
Magnetic field at $3.00 mm$ outside the wire is $167 渭罢$
Magnetic field at $2.20 cm$ outside the wire is $22.7 渭罢$
Magnetic field due to $i_{d}$ at $1.00 mm$ inside the gap is $1.25 渭 T$
Magnetic field due to $i_{d}$ at $3.00$ mm inside the gap is $3.75 渭罢$
Magnetic field due to $i_{d}$ at $2.20 cm$ outside the gap is $22.7 渭罢$
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 by step solution

Listing the given quantities

Charged current is $i = 2.50 A$

Wire radius is $R = 1.50 mm$

Plate radius is $r = 2.00 cm$

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:

$B = \frac{渭_{0} ir}{2 {蟺搁}^{2}}$

$B = \frac{渭_{0} i}{2 蟺谤}$

(a) Calculations of the Magnetic field at 1.00 mm inside the wire

Using the formula given below,

$\begin{matrix} B = \frac{渭_{0} ir}{2 {蟺搁}^{2}} \\ = \frac{4 蟺脳 10^{鈭� 7} 脳 2.50 脳 2.50 脳 10^{鈭� 2}}{2 蟺脳 {(1.50 脳 10^{鈭� 3})}^{2}} \\ = 2.22 脳 10^{- 4} T \\ = 222 渭罢 \end{matrix}$

Magnetic field at $1.00 mm$ inside the wire is $222 渭罢$

(b) Calculations of the magnetic field at 3.0 mm outside the wire

Usethefollowing formula:

$\begin{matrix} B = \frac{渭_{0} i}{2 蟺谤} \\ = \frac{4 蟺脳 10^{鈭� 7} 脳 2.50}{2 蟺脳 3 脳 10^{鈭� 3}} \\ = 1.667 脳 10^{- 4} T \\ = 167 渭罢 \end{matrix}$

Magnetic field at $3.00 mm$ outside the wire is $167 渭罢$

(c) Calculations of the Magnetic field at 2.20cm outside the wire

$\begin{matrix} B = \frac{渭_{0} i}{2 蟺谤} \\ = \frac{4 蟺脳 10^{鈭� 7} 脳 2.50}{2 蟺脳 2.20 脳 10^{鈭� 2}} \\ = 2.27 脳 10^{- 5} T \\ = 22.7 脳 10^{- 6} T \\ = 22.7 渭罢 \end{matrix}$

Magnetic field at $2.20 cm$ outside the wire is $22.7 渭罢$

(d) Calculations of the Magnetic field due to id at 1.00 mm inside the gap

Use the following formula:

$\begin{matrix} B = \frac{渭_{0} i_{d} r}{2 {蟺搁}^{2}} \\ = \frac{4 蟺脳 10^{鈭� 7} 脳 2.50 脳 1 脳 10^{鈭� 3}}{2 蟺脳 {(2 脳 10^{鈭� 2})}^{2}} \\ = 1.25 脳 10^{- 6} T \\ = 1.25 渭罢 \end{matrix}$

Magnetic field due to $i_{d}$ at $1.00 mm$ inside the gap is $1.25 渭罢$

(e) Calculations of the magnetic field due to id at 3.0mm inside the gap

Use the following formula:

$\begin{matrix} B = \frac{渭_{0} ir}{2 {蟺搁}^{2}} \\ = \frac{4 蟺脳 10^{鈭� 7} 脳 2.50 脳 3 脳 10^{鈭� 3}}{2 蟺脳 {(2 脳 10^{鈭� 2})}^{2}} \\ = 3.75 脳 10^{- 6} T \\ = 3.75 渭罢 \end{matrix}$

Magnetic field due to $i_{d}$ at $3.00$ mm inside the gap is $3.75 渭罢$

(f) Calculations of the magnetic field due to id at 2.20 cmoutside the gap

Use the following formula:

$\begin{matrix} B = \frac{渭_{0} ir}{2 {蟺搁}^{2}} \\ = \frac{4 蟺脳 10^{鈭� 7} 脳 2.50}{2 蟺脳 (2.2 脳 10^{鈭� 2})} \\ = 2.27 脳 10^{- 5} T \\ = 22.7 渭罢 \end{matrix}$

Magnetic field due to role="math" localid="1662983978538" $i_{d}$ at $2.20 c m$ outside the gap is $22.7 渭罢$

(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.