Complete Question
A 8.0 A current is charging a 1.0 -cm-diameter parallel-plate capacitor. What is the magnetic field strength at a point 2.5 mm radially from the center of the wire leading to the capacitor?
Answer:
The magnetic field is [tex]B = 6.4*10^{-4} \ T[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 8.0 \ A[/tex]
The diameter is [tex]d = \ cm = \frac{1}{100} = 0.01 m[/tex]
The position considered is [tex]d = 2.5 \ mm = 0.0025 \ m[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * I }{ 2 \pi d}[/tex]
Here [tex]\mu_o[/tex] is permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} \ N/A^2[/tex]
So
[tex]B = \frac{ 4\pi * 10^{-7} * 8 }{ 2 \pi * 0.0025 }[/tex]
=> [tex]B = \frac{ 2 * 10^{-7} * 8 }{ * 0.0025 }[/tex]
=> [tex]B = 6.4*10^{-4} \ T[/tex]
