Answer:9.34 A/s
Explanation:
Given
radius of solenoid [tex]R=1.06 m[/tex]
Emf induced [tex]E=8.50\times 10^{-6} V/m[/tex]
no of turns per meter n=450
we know Induced EMF is given by
[tex]\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A[/tex]
Magnetic Field is given by
[tex]B=\mu _0ni[/tex]
thus [tex]\frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}[/tex]
Area of cross-section
[tex]A=\pi R^2[/tex] where
solving integration we get
[tex]E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2[/tex]
where r=distance from axis
R=radius of Solenoid
[tex]\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}[/tex]
[tex]\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}[/tex]
[tex]\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s[/tex]
