Question

What is the induced magnetic field in-between the plates of the capacitor?

Problem 10 Objective #4: A parallel plate capacitor is made from two identical circular disks, each with a radius of 10 m. The space in between the plates of the capacitor is filled with a material with Ke 2 and km-2. The capacitor is charging at a rate of 1 C/s. What is the magnitude of the induced magnetic field in-between the plates of the capacitor a distance of 5 m from the axis of symmetry of the capacitor? :10 m m s e, Knm front view side view Answers: (a) 10 nT (b) 20 nT (c) 40 nT (d) 80 nT (e) 160 nT

Answer 1

The electric field in between the plates or radius R with cross sectional area A is given by
   $E=\frac{\sigma}{\epsilon}=\frac{Q}{\epsilon A}$
where, Q is the charge on the plates at a instant.
So, the displacement current in between the plates is given by
   
1 OtA dt

And so, the magnetic field at a distance r from the central axis of the plates is
  
Β.dl= μ | Ja.ds

  $\Rightarrow B.2\pi r=\mu J_{d}.\pi r^2$
  $\Rightarrow B=\frac{\mu r}{2} J_{d}$
   
2.4 dt


Rkmplor do 24 dt

  And as the radius of the plate is R, so, we get
   
Rkmor do 2TR2 dt

Now putting the given values,
   
de 4-1 C/s. dt μ0 = 4π × 10-7 T.rn/.A km =2. R=10m. r=5m.

we get
   
2 × 4π × 10-1 × 5 2π × 102

   $\Rightarrow B=20\times 10^{-9}~T$
  
B=20 nT

So, option (b) is correct.