Answer:
[tex]E=1.028x10^{-4} V[/tex]
Explanation:
[tex]N=500\\n=10[/tex]
[tex]d=25cm*\frac{1m}{100cm}=0.25m[/tex]
[tex]I=5A\\t=0.6s[/tex]
[tex]u_{o}=4\pi x10^{-7}\frac{T*m}{A}[/tex]
So the field create inside the coil can be find:
[tex]\beta =\frac{u_{o}*N*i}{L}[/tex]
The rate of tha change of B field is:
Δβ/Δt=[tex]\frac{u_{o}*N*\frac{di}{dt}}{L}[/tex]
[tex]\frac{di}{dt}=\frac{5A}{0.6s}=8.333 \frac{A}{s}[/tex]
Δβ/Δt=[tex]\frac{4\pi x10^{-7}\frac{T*m}{A} *500*8.33\frac{A}{s}}{0.25m}[/tex]
Δβ/Δt=[tex]20.935x10^{-3} \frac{T}{s}[/tex]
Induce E caused by tha changing B is:
E=[(Δβ/Δt)*A*n]
[tex]r=2.5cm=r=0.025m\frac{0.025m}{2}=0.0125m\\A=\pi *r^{2}=\pi*(1.25x10^{-2}m)^{2}\\A=4.91x10^{-4} m^{2}\\E=(2.09x10^{-2}\frac{T}{s}*4.91x10^{-4}m^{2}*10) \\[/tex]
E=1.028[tex]x10^{-4}[/tex]V
