physics-class-12-unit-08-chapter-04-problems-in-electromagnetics-magnetic-fields-em-waves-l-4-4-8cxrdikqgb4

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Solution-1 </primary>
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- $\oint \vec{B}.d \vec{A}=0$

- $\oint \vec{B}.d \vec{A} = \int_{S_1} \vec{B}.\vec{dA} + \int_{S_2} \vec{B}.\vec{dA} + \int_{S_3} \vec{B}.\vec{dA}+\int_{S_4} \vec{B}.dA + \int_{S_5} \vec{B}.\vec{dA} + \int_{S_6} \vec{B}.dA$

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<footer style = "font-size: 18px">Problems in Electromagnetics $\rightarrow$ Probelm-1 $\rightarrow$ <span style = "color:blue"> Solution-1 </span> $\rightarrow$ Vector Fields $\rightarrow$ Vector Fields </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Vector Fields </primary>
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- $\int_{S_1} \vec{B} d \vec{A}$

- $d \vec{A}=dydz \hat{i}$

- $vec{B}=kx \hat{i} + ky \hat{j}$

- $\vec{B}.d \vec{A}=(kx \hat{i} + ky \hat{j}). \hat{i} dydz$

- $=kadydz$

- $\int_{S_1} \vec{B}.d \vec{A}=\int_{S_1} kadydz=ka \int_{S_1} dydz$

- $=ka^3$

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<footer style = "font-size: 18px">Probelm-1 $\rightarrow$ Solution-1 $\rightarrow$ <span style = "color:blue"> Vector Fields </span> $\rightarrow$ Vector Fields $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- Calculate the magnetic field produced by a finite current element as shown

- $d \vec{B}=\frac{\mu_0}{4 \pi} I \frac{\vec{dl} \times \vec{s}}{s^3}$

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<footer style = "font-size: 18px">Vector Fields $\rightarrow$ Vector Fields $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution-2 $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Solution-2 </primary>
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- $\vec{dl} \times \vec{s}=dl s \sin \theta$

- $\sin \theta = \cos \alpha$

- $\vec{dl} \times \vec{s}=dl s \cos \alpha$

- $s^2 = r^2 + z^2$

- $|d \vec{B}|=\frac{\mu_0 I}{4 \pi} \frac{dl s \cos \alpha}{s (r^2 + z^2)}$

- $=\frac{\mu_0 I}{4 \pi}\frac{\cos \alpha dz}{r^2 + z^2}$

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<footer style = "font-size: 18px">Vector Fields $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution-2 </span> $\rightarrow$ Magnetic Field $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Magnetic Field </primary>
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- $|\vec{B}|=\frac{\mu_0 I}{4 \pi} \int_{z_1}^{z-2} \frac{dz \cos \alpha}{(r^2=z^2)}$

- $z=r \tan \alpha$

- $dz=r \sin^2 \alpha d \alpha$

- $r^2 + z^2 = r^2 (1+\tan^2 \alpha)=r^2 \sec^2 \alpha$

- $|\vec{B}|=\frac{\mu_0I}{4 \pi} \int \frac{r \sec^2 \alpha d \alpha \cos \alpha}{r^2 \sec^2 \alpha}$

- $=\frac{\mu_0I}{4 \pi r} \int \cos \alpha d \alpha$

- $=\frac{\mu_0I}{4 \pi r}(\sin \alpha_2 - \sin \alpha_1)$

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution-2 $\rightarrow$ <span style = "color:blue"> Magnetic Field </span> $\rightarrow$ Magnetic Field $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- A length L of wire carrying a current I is to be went into a circle or a square each of one turn. In which case will a magnetic field at the centre be greater?

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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Magnetic Field $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution-3 $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Solution-3 </primary>
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- $|\vec{B}|=[\frac{\mu_0I}{4 \pi r}[\frac{z^2}{\sqrt{z_2^2 + r^2}}-\frac{z_1}{\sqrt{z_1^2+r^2}}]$

- $z_1 = -L/8$

- $z_2 =+L/8$

- $r=L/8$

- $\frac{z_2}{\sqrt{z_2^2 + r^2}}=\frac{L/8}{L/8 \sqrt{2}}=\frac{1}{\sqrt{2}}$

- $\frac{z_1}{\sqrt{z_1^2 + r^2}}=\frac{-L/8}{L/8 \sqrt{2}}=-\frac{1}{\sqrt{2}}$

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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution-3 </span> $\rightarrow$ Magnetic Field $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Magnetic Field </primary>
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- Magnetic field due to one side of the square

- $=\frac{\mu_0 I}{4 \pi \frac{L}{8}}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})$

- $=\frac{2 \mu_0 I}{\pi L}\sqrt{2}$

- Total magnetic fields $=4 \times \frac{2 \mu_0 I}{\pi L}\sqrt{2}$

- $B_s = \frac{8 \sqrt{2} \mu_0 I}{\pi L}$

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution-3 $\rightarrow$ <span style = "color:blue"> Magnetic Field </span> $\rightarrow$ Magnetic Field $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Magnetic Field </primary>
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- $Circle$

- $L=2 \pi R$

- $R=\frac{L}{2 \pi}$

- $dB=\frac{\mu_0 I}{4 \pi}.\frac{Rd Q}{R^2}$

- $=\frac{\mu_0 I d Q}{4 \pi R}$

- $B_c=\frac{\mu_0I}{4 \pi R} \int_0^{2 \pi} dQ=\frac{\mu_0 I}{2R}=\frac{\mu_0 I \pi}{L}$

- $B_c=\frac{\mu_0I \pi}{L}$

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<footer style = "font-size: 18px">Solution-3 $\rightarrow$ Magnetic Field $\rightarrow$ <span style = "color:blue"> Magnetic Field </span> $\rightarrow$ Magnetic Field $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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-  A finite wire is used to curved six turns around an insulating sphere of radius a such that each turn mades an angle of $30^o$ with the adjacent turns and that all the turns intersect at diagramatically opposite points on the surface of the sphere. If a current I is passed through that turns, Find the magnitude of the magnetic field at the centre of the sphere.
 
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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Magnetic Field $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution-4 $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Magnetic Field </primary>
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- Component of magnetic field along the verticle direction

- $B_v=\frac{\mu_0 I}{2a}+\frac{\mu_0 I}{2a}\cos60^o +\frac{\mu_0 I}{2a}\cos60^o +\frac{\mu_0 I}{2a}\cos90^o+\frac{\mu_0 I}{2a}\cos120^o +\frac{\mu_0 I}{2a}\cos150^o$

- $=\frac{\mu_0 I}{2a}[1+ \frac{\sqrt{2}}{2}+ \frac{1}{2}+ 0- \frac{1}{2}- \frac{\sqrt{3}}{2}]$

- $=\frac{\mu_0 I}{2a}$

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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution-4 $\rightarrow$ <span style = "color:blue"> Magnetic Field </span> $\rightarrow$ Magnetic Field $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- A long solid conducting cylinder of radius R has a cyclindrical hole of radius a drilled onto such that the axis of the hole is parallel to the axis of the cylinder. If b is the distance between the two axis and a current I is passing through the remaining solid cylinder show that the magnetic field is constant through out the hole.

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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Magnetic Field $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution-5 $\rightarrow$ Magnetic Field </footer>

<h3 id="success-stylefont-size25px-problems-in-electromagnetics-magnetic-fields-em-waves-l-4-success"><Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-5-primary"><primary style="font-size:24px"> Solution-5 </primary></h3>
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<p>Current density</p>
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<p>$I=\frac{I}{\pi(R^2-a^2)}$</p>
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<p>Magnetic field due to solid conduction (without the hole) of radius R</p>
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<p>$\oint \vec{B}. \vec{dl}= \mu_0 I_{enc}$</p>
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<p>$2 \pi r B = \mu_0 J. \pi r^2$</p>
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<p>$B=\frac{\mu_0 Jr}{2}; \vec{B}=\frac{\mu_0 Jr}{2}\hat{m}$</p>
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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution-5 </span> $\rightarrow$ Magnetic Field $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Magnetic Field </primary>
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- Magnetic field due to solid cyclinder of radius a at a distance from the centre

- $\oint \vec{B}. \vec{dl}=\mu_0 I_{enc}$

- $2 \pi 5 B=\mu_0 J. \pi 5^2$

- $B=\frac{\mu_0 J5}{2}$

- $\vec{B_2}=\frac{\mu_0 J5}{2}\hat{n_2}$

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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution-5 $\rightarrow$ <span style = "color:blue"> Magnetic Field </span> $\rightarrow$ Magnetic Field $\rightarrow$ Magnetic Field </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Magnetic Field </primary>
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- Total field

- $\vec{B}=\vec{B}_1 - \vec{B}_2$

- $=\frac{\mu_0 J}{2}(r \hat{n}_1 - 5 \hat{n}_2)$

- $=\frac{\mu_0 J}{2} [r(- \sin \theta \hat{i} + \cos \theta \hat{j})+5(\sin \phi \hat{i} + \cos \phi \hat{j})]$

- $=\frac{\mu_0 J}{2}[\hat{i} (-r \sin \theta + 5 \sin \phi) + \hat{j}(r \cos \theta + 5 \cos \phi)]$

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<footer style = "font-size: 18px">Solution-5 $\rightarrow$ Magnetic Field $\rightarrow$ <span style = "color:blue"> Magnetic Field </span> $\rightarrow$ Magnetic Field $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Problems In Electromagnetics Magnetic Fields Em Waves L-4 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- Electric field of a plane electromagnetic wave propagating in force space is given by

- $\vec{E}=(10 \hat{i} + 15 \hat{j}) \sin [4 \pi \times 10^5(ct - z)]^{v/m}$

- c : speed of light in free space

- z : meter

- a) What is the wavelength of the wave

- b) Calculate the corresponding $\vec{B}$ field

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<footer style = "font-size: 18px">Magnetic Field $\rightarrow$ Magnetic Field $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Solution-6 $\rightarrow$ Thank You </footer>