Question

Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.

Answer 1

a

`\phi = 1.78 *10^(-7) \ Weber`

b

`L = 1.183 *10^(-7) \ H`

Step-by-step explanation:

From the question we are told that

The radius is
`r = 6 \ cm = (6)/(100) = 0.06 \ m`

The current it carries is
`I = 1.50 \ A`

The magnetic flux of the coil is mathematically represented as

`\phi = B * A`

Where B is the magnetic field which is mathematically represented as

`B = (\mu_o * I)/(2 * r)`

Where
`\mu_o` is the magnetic field with a constant value
`\mu_o = 4\pi * 10^(-7) N/A^2`

substituting value

`B = (4\pi * 10^(-7) * 1.50 )/(2 * 0.06)`

`B = 1.571 *10^(-5) \ T`

The area A is mathematically evaluated as

`A = \pi r ^2`

substituting values

`A = 3.142 * (0.06)^2`

`A = 0.0113 m^2`

the magnetic flux is mathematically evaluated as

`\phi = 1.571 *10^(-5) * 0.0113`

`\phi = 1.78 *10^(-7) \ Weber`

The self-inductance is evaluated as

`L = (\phi )/(I)`

substituting values

`L = (1.78 *10^(-7) )/(1.50 )`

`L = 1.183 *10^(-7) \ H`