Answer to Question #191225 in Electricity and Magnetism for Hanatayuji

Question

Given dielektric water constant is 80, if mean of momen dipole molecular magnetic water is 6,2×10^-3, what is polarization and magnetic field

Answer 1

Solution

To be given in ques tion

$\epsilon_r=80$

Water dipole moment p=$6.2\times10^{-30} c-m$

$N_A=6.625\times10^{23} mol^{-1}$

Water polerization

$P=Np\rightarrow equation (1)$

Put value

$P=6.625\times10^{23}\times6.2\times10^{-30}$

P=$4.10\times10^{-6}C/m^2$

We know that

$E=\frac{P}{\epsilon_0 (\epsilon_r -1)}\rightarrow(2)$

Eqution (2)put value

$E=\frac{4.10\times10^{-6}}{8.85\times10^{-12} (80 -1)}$

$E=5.87\times10^{3}V/m$

We know that

$E=Bc$

$B=\frac{E}{c}\rightarrow(3)$

$B=\frac{5.87\times10^3}{3\times10^8}$

$B=1.95\times10^{-5} T$