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Yogita Ingle 6 years, 5 months ago
The direction of the current induced in the conductor due to this induced EMF is such that it opposes the change that produced it in the first place.In other words, the induced EMF has a polarity which opposes the change in the magnetic flux through the conducting loop. Hence, induced EMF is also called Back EMF.
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Yogita Ingle 6 years, 5 months ago
Principle of a capacitor : A capacitor works on the principle that the capacitance of a conductor increase appreciably when an earthed conductor is brought near it. Thus a capacitor has two plates separated by a distance having equal and opposite charges.
Parallel Plate Capacitor
A parallel plate capacitor is a capacitor with 2 large plane parallel conducting plates separated by a small distance.
- Electric field inside the capacitor has a direction from positive to negative plate.
- For very small‘d’, the electric field is considered as uniform. For large‘d’, the electric field is non-uniform and it bends around the corners of the plate which is called fringing of the field.
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Sia ? 6 years, 5 months ago
Magnetic permeability of a material is the ability of a material to support the formation of a magnetic field inside itself.
Magnetic susceptibility is the measure of magnetic properties of a material which indicates whether the material is attracted or repelled from an external magnetic field.
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Sia ? 6 years, 5 months ago
Mutual Inductance is the basic operating principal of the transformer, motors, generators and any other electrical component that interacts with another magnetic field. Then we can define mutual induction as the current flowing in one coil that induces a voltage in an adjacent coil.
derive the expression for the mutual inductance of two long coaxial solenoids of same length l having radii r1 and r2 (r2>r1 and l>>r2). Suppose a current i is passed through the inner solenoid . A magnetic field is produce inside whereas field outside it is zero.
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Microwaves are considered suitable for radar systems used in aircraft navigation because they have a short wavelength range (10-3 m to 0.3 m), which makes them suitable for long range communication.
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Sia ? 6 years, 5 months ago
{tex}\rho=\frac{q}{V}{/tex}and {tex}\sigma=\frac{q}{A}{/tex} respectively, where,
<script type="math/tex" id="MathJax-Element-3">\qquad q</script> is the charge, <nobr aria-hidden="true">V</nobr> is the volume and <nobr aria-hidden="true">A</nobr> is the area.
Let the radii of the two spheres be <nobr aria-hidden="true">r1 </nobr>and <nobr aria-hidden="true">r2 </nobr> respectively.
It is given that the charge on both the spheres is equal.
{tex}\Rightarrow \quad \rho_{1}=\frac{q}{\frac{4}{3} \pi r_{1}^{3}} \text { and } \rho_{2}=\frac{q}{\frac{4}{3} \pi r_{2}^{2}}, \text { and }{/tex}
{tex}\sigma_{1}=\frac{q}{4 \pi r_{1}^{2}} \text { and } \sigma_{2}=\frac{q}{4 \pi r_{1}^{2}}{/tex}
<script type="math/tex" id="MathJax-Element-11">\sigma_2 = \frac{q}{4\pi r_1^2}.</script>
It is given that the ratio of the volume density is <nobr aria-hidden="true">8:64.</nobr>
{tex}\Rightarrow \quad \frac{\rho_{1}}{\rho_{2}}=\frac{8}{64} \quad \Rightarrow \quad \frac{\frac{3 q}{4 \pi_{1}^{3}}}{\frac{3 q}{4 m_{2}^{3}}}=\frac{8}{64}{/tex}
{tex}\Rightarrow \quad\left(\frac{n}{n}\right)^{3}=\frac{8}{\sqrt{4}} \quad \Rightarrow \quad \frac{n}{n}=\frac{1}{2} \quad \Rightarrow \quad\left(\frac{n}{n}\right)^{2}=\frac{1}{4}{/tex}
{tex}\Rightarrow{/tex} The ratio of the surface density of charge is 1 : 4
<script type="math/tex" id="MathJax-Element-15">\Rightarrow \qquad \frac{\sigma_1}{\sigma_2} = \frac{\frac{q}{4\pi r_1^2}}{\frac{q}{4\pi r_2^2}} = \left(\frac{r_2}{r_1}\right)^2 = \frac{1}{4}.</script>
</div> </div> </div> </div> </div> </div> </div> </div> </div> </div> </div> </div>Posted by Shubhdeep Paul 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
As we know
{tex}I = \frac{E}{{R + r}}{/tex}
Now, {tex}0.5 = \frac{E}{{12 + r}}{/tex} and {tex}0.25 = \frac{E}{{25 + r}}{/tex}
- Dividing, {tex}\frac{{0.5}}{{0.25}} = \frac{{25 + r}}{{12 + r}}{/tex}
or, {tex}2 = \frac{{25 + r}}{{12 + r}}{/tex} or {tex}r = 1\Omega{/tex} - {tex}0.5 = \frac{E}{{12 + 1}} \Rightarrow{/tex} E = 6.5 V
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Khuskaran Sidhu 6 years, 5 months ago
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