Q74P A parallel-plate capacitor with ... [FREE SOLUTION]

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Chapter 32: Q74P (page 971)

A parallel-plate capacitor with circular plates is being charged. Consider a circular loop centered on the central axis and located between the plates. If the loop radius of $3.00cm$ is greater than the plate radius, what is the displacement current between the plates when the magnetic field along the loop has magnitude $2.00渭罢$ ?

Short Answer

Expert verified

The displacement current between the parallel plate capacitor is $i_{d} = 0.300A$ .

Step by step solution

Listing the given quantities:

The radius of the loop is, $r = {3.00肠尘=3.00脳10}^{-2} m$

The magnitude of magnetic field is, $B = 2.00鈥坝糡= 2.00 脳 10^{- 6} T$

Understanding the concepts of displacement current:

You can use the concept of the displacement current and the expression of the magnetic field outside a circular capacitor.

Formula:

$B = \frac{渭_{0} i_{d}}{2 蟺 r}$

Here, $渭_{0}$ is the permeability of free space having a value $4 蟺脳 10^{- 7} \frac{T 鈰� m}{A}$ , $i_{d}$ is the displacement current, and $r$ is the distance.

Calculations of the displacement current between the parallel plate capacitor:

The expression of the magnetic field outside a circular capacitor in terms of the displacement current is,

$\begin{array}{rcl} B & = & \frac{渭_{0} i_{d}}{2 蟺 r} \\ i_{d} & = & \frac{2 蟺 r B}{渭_{0}} \\ = & \frac{2 脳 3.14 脳 3.00 脳 10^{- 2} m 脳 2.00 脳 10^{- 6} A}{4 蟺脳 10^{- 7} \frac{T 鈰� m}{A}} \\ = & 0.300 A \end{array}$