Answer:
The current is [tex]I =0.2mA[/tex]
Explanation:
From the question we are told that
The first radius is [tex]R_1 = 5cm = \frac{5}{100} = 0.05cm[/tex]
The number of turns is [tex]N = 17 \ turn/cm[/tex]
The current rate is [tex]\frac{dI}{dt} = 5 A/s[/tex]
The second radius is [tex]R_2 = 7cm = \frac{7}{100} = 0.07m[/tex]
The resistance is [tex]r = 5 \Omega[/tex]
Generally the magnetic flux induced in the solenoid is mathematically represented as
[tex]\O = B A[/tex]
Where is the magnetic field mathematically represented as
[tex]B = N \mu_o I[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a value of [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
and A is the area mathematically represented as
[tex]A = \pi (R_2 - R_1)^2[/tex]
So
[tex]\O = N \mu I * \pi R^2[/tex]
Substituting values
[tex]\O = 17 * 4\pi *10^{-7} * \pi (7-5)^2I[/tex]
[tex]\O = 2.68*10^{-4}I[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- |\frac{d\O}{dt}|[/tex]
[tex]\epsilon = 2.68*10^{-4 } \frac{dI}{dt}[/tex]
substituting values
[tex]\epsilon =2.68 *10^{-4} * 5[/tex]
[tex]=1.3 *10^{-3} V[/tex]
From Ohm law
[tex]I = \frac{\epsilon }{r}[/tex]
Substituting values
[tex]I = \frac{1.3*0^{-3}}{5}[/tex]
[tex]I =0.2mA[/tex]
