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The ratio of electric field vectors  \vec{E}  and magnetic field vector \vec{H}  ie. \left(\frac{\bar{E}}{H}\right)  has the dimensions of -

Option: 1

Resistance


Option: 2

Inductance


Option: 3

capacitance
 


Option: 4

Product of induectance and capacitance.


Answers (1)

we know that,

H=\frac{1}{\mu_0 c} \vec{E}
\left|\frac{E}{H}\right|=\left|\frac{E_0}{H_0}\right|=\mu_0 \mathrm{c} \\

E=\text { Volt } / m \\
 H=Amphere - turn/m 
\text { ratio, } \quad z_0=\frac{v_{\text {olt }} / \mathrm{m}}{\text { Amp.tum } / \mathrm{m}}=
v_{\text {olt }} & =\text { Resistance } \\
\frac{v_{\text {olt }} }{\text { Amp }}= resistance

               =\Omega
 

Note-from Ohm's law V=I K

\operatorname{Volt} / \text { Amp. } \longrightarrow \frac{V}{I}=k

Posted by

Ramraj Saini

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