Question

Calculate the magnetic field strength with a distance of 1.5 cm from the axis of a 2.0 cm radius capacitor with a 1.0 mm spacing that is being charged by a 1.5 A current.

Answer 1

Answer:

Magnetic Field Strength = $1.126$x$10^{-5}$ T

Explanation:

We can calculate the magnetic field strength with the formula

$B =\frac{ (\mu a I)}{ 2A}$

Where A = area of the circular magnetic field = π$r^{2}$

a = radial distance = 1.5 cm

r = radius of the capacitor plates = 2 cm

μο= 4πx$10^{-7} H/m$

Current I = 1.5 A

What remains is to plug in the values into the equation for the magnetic field strength and get the solution.

$B= \frac{(4\pi \times 10^{-7}\times (1.5\times10^{-2})\times 1.5) }{2\times \pi\times (2\times10^{-2})^{2} }$

$B= \frac{2.83\times10^{-8} }{2.513\times10^{-3} }=1.126\times 10^{-5}$

= $1.126$x$10^{-5}$ T

Magnetic Field Strength = $1.126$x$10^{-5}$ T