Answer:
1.8 mV
Explanation:
The total number of turns in the solenoid is
[tex]N=nL = (6500)(9.10\cdot 10^{-2})=592[/tex]
where
n = 6500 is the number of turns per meter of length
[tex]L=9.10\cdot 10^{-2} m[/tex] is the length of the solenoid
The flux linkage through the solenoid is given by
[tex]\Phi = NBA[/tex]
where
B is the strength of the magnetic field
[tex]A=5.0\cdot 10^{-5}m^2[/tex] is the cross-sectional area of the solenoid
The strength of the field in the solenoid is given by
[tex]B=\mu_0 n I[/tex]
where I is the current.
At the beginning, I = 0, so the field is
[tex]B_i=0[/tex]
And the flux linkage is
[tex]\Phi_i = NAB_i = 0[/tex]
Later, the current is I = 1.5 A, so the field in the solenoid is
[tex]B_f = (4\pi \cdot 10^{-7})(6500)(1.5)=0.0122 T[/tex]
So, the flux linkage is
[tex]\Phi_f = (592)(5.0\cdot 10^{-5})(0.0122)=3.61\cdot 10^{-4}Wb[/tex]
So, the change in flux linkage is
[tex]\Delta \Phi = \Phi_f - \Phi_i = 3.61\cdot 10^{-4} Wb[/tex]
And therefore, the emf induced in the solenoid is (in magnitude)
[tex]\epsilon=\frac{\Delta \Phi}{\Delta t}=\frac{3.61\cdot 10^{-4} Wb}{0.20 s}=1.8\cdot 10^{-3} V = 1.8 mV[/tex]
