Question

For the situation described in figure , the magnetic field changes with time according to $B=(2.00t^3-4.00t^2+0.8)T$ and $r_2=2R=5.0cm$
a) Calculate the force on an electron located at$P_2$ at t=2.00 s
b) What are the magnitude and direction of the electric field at $P_1$ when t=3.00 s and $r_1=0.02m,$

Answer 1

Flux $\varphi =\frac{d(B.ds)}{dt}$

where $dB$=area vector and $\overrightarrow{B}$=magnetic field

emf induced across loop of ares $ds$ is,

$e=-\frac{dQ}{dt}$

(a) $e=-A\frac{dB}{dt}$ where $A$ is area of loop

$=-A({6t}^2-8t)att=2s$

$e=-\pi R^2(6\times 4-8\times 2)$ (Area=$\pi R^2$)

$=-\pi R^2(24-16)$

$=-8\pi {\left(2.5\right)}^2\times {10}^{-4}$

$=-50\times {10}^{-4}$

Voltage difference =$\left|e\right|=50\pi \times {10}^{-4}$ between $c$ and $P_2$.

Electric field $\overrightarrow{E}$ is $E$, $V$=$\oint E.dt$

$\oint dl=2\pi r_2=2\pi \times 5\times {10}^{-2}=10\pi \times {10}^{-2}$

$E=\frac{V}{10\pi \times {10}^{-2}}=\frac{50\pi \times {10}^{-4}}{10\pi \times {10}^{-2}}=5\times {10}^{-2}\frac{v}{m}.$

Force on electron =$qE$

$F_e$=$1.6\times {10}^{-19}\times 5\times {10}^{-2}$

$F_e$=$8\times {10}^{-21}$.N

(b) $e$ at $t=3s$ at point $P_1$

$e=-\left(\pi {r_1}^2\right)(6\times 3^2-8\times 3)$

$=-\pi {\left(0.02\right)}^2(56-24)$

$=-4\pi \times {10}^{-4}\times 32$

$=-128\pi \times {10}^{-4}$

$E_{p1}=\frac{\left|e\right|}{\oint dl}=\frac{128\pi \times {10}^{-4}}{2\pi \times r_1}$

$=\frac{128\pi \times {10}^{-4}}{2\pi \times 2\times {10}^{-2}}=32\times {10}^{-2}$

$E_{P1}$=$0.32\frac{v}{m}$