# Problem: A 5.0-cm-diameter coil has 20 turns and a resistanceof 0.50 ohms. A magnetic field perpendicular to the coil is B =0.02t + 0.010t2, where B is in tesla and t is in seconds. Find an expression for the induced current I(t)as a function of time.

###### FREE Expert Solution

Current:

$\overline{){\mathbf{i}}{\mathbf{=}}\frac{\mathbf{N}\mathbf{\pi }{\mathbf{r}}^{\mathbf{2}}}{\mathbf{R}}{\mathbf{\left(}}\frac{\mathbf{d}\mathbf{B}}{\mathbf{d}\mathbf{t}}{\mathbf{\right)}}}$

N is the number of turns, r is the radius of the coil, R is the resistance, and dB/dt is the derivative of the magnetic field with respect to time.

$\frac{\mathbf{d}\mathbf{B}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{\left(}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{t}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{010}{\mathbf{t}}^{\mathbf{2}}\mathbf{\right)}$

90% (475 ratings) ###### Problem Details

A 5.0-cm-diameter coil has 20 turns and a resistanceof 0.50 ohms. A magnetic field perpendicular to the coil is B =0.02t + 0.010t2, where B is in tesla and t is in seconds.

Find an expression for the induced current I(t)as a function of time.