Question

A long, thin solenoid has 450 turns per meter and a radius of 1.04 cm . The current in the solenoid is increasing at a uniform rate didt. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.55 cm from its axis is 8.00×10?6 V/m .

a)Calculate didt.

Answer 1

Given

E = 8x10^-6 V/m

n = 450 m^-1

$\oint Edl=-\frac{\mathrm{d\phi } }{\mathrm{d} t}$

$\varepsilon =\frac{d\Phi _B}{dt}=\frac{d}{dt}(BA)=\frac{\mathrm{d} }{\mathrm{d} t}(\mu _onIA)=\mu _onA\frac{\mathrm{dI} }{\mathrm{d} t}=\oint Edl$

$\frac{\mathrm{dI} }{\mathrm{d} t}=\frac{E.2\pi r}{\mu _onA}=\frac{(8\times10^{-6}V/m)(2\pi )(0.0355m)}{(4\pi \times10^{-7})(450m^{-1})\pi (1.04\times10^{-2}m)^2}=9.28A/s$