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

Answer:

$E = 2 \times 10^{-3} V$

Explanation:

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

So here we can

$EMF = \frac{d\phi}{dt}$

now we have

$EMF = \pi r^2\frac{dB}{dt}$

now we also know that induced EMF is given by

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

$E (2\pi r) = \pi r^2\frac{dB}{dt}$

$E = \frac{r}{2}(\frac{dB}{dt})$

now plug in all values in it

$E = \frac{0.0133}{2}(0.12 t)$

$E = 8 \times 10^{-4} (2.50) = 2 \times 10^{-3} V/m$