Question

A magnetic field directed along the x-axis changes with time according to B (0.06t2+2.25) T, where t is in seconds. The field is confined to a circular beam of radius 2.00 cm. What is the magnitude of the electric field at a point 1.33 cm measured perpendicular from the x-axis when t 2.50 s? N/m

Answer 1

`E = 2 * 10^(-3) V`

Step-by-step explanation:

As we know that rate of change in flux will induce EMF

So here we can

`EMF = (d\phi)/(dt)`

now we have

`EMF = \pi r^2(dB)/(dt)`

now we also know that induced EMF is given by

`\int E. dL = \pi r^2\frac{dB}[dt}`

`E (2\pi r) = \pi r^2(dB)/(dt)`

`E = (r)/(2)((dB)/(dt))`

now plug in all values in it

`E = (0.0133)/(2)(0.12 t)`

`E = 8 * 10^(-4) (2.50) = 2 * 10^(-3) V/m`