Answer:
a
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
b
[tex]L = 1.183 *10^{-7} \ H[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 6 \ cm = \frac{6}{100} = 0.06 \ m[/tex]
The current it carries is [tex]I = 1.50 \ A[/tex]
The magnetic flux of the coil is mathematically represented as
[tex]\phi = B * A[/tex]
Where B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * I}{2 * r}[/tex]
Where [tex]\mu_o[/tex] is the magnetic field with a constant value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = \frac{4\pi * 10^{-7} * 1.50 }{2 * 0.06}[/tex]
[tex]B = 1.571 *10^{-5} \ T[/tex]
The area A is mathematically evaluated as
[tex]A = \pi r ^2[/tex]
substituting values
[tex]A = 3.142 * (0.06)^2[/tex]
[tex]A = 0.0113 m^2[/tex]
the magnetic flux is mathematically evaluated as
[tex]\phi = 1.571 *10^{-5} * 0.0113[/tex]
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
The self-inductance is evaluated as
[tex]L = \frac{\phi }{I}[/tex]
substituting values
[tex]L = \frac{1.78 *10^{-7} }{1.50 }[/tex]
[tex]L = 1.183 *10^{-7} \ H[/tex]
