Consider a vector field given by

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

Where k is a constant can the vector field represent the magnetic fields?

$\oint \vec{B}.d \vec{A}=0$

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

$\int_{S_1} \vec{B} d \vec{A}$

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

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

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

$=kadydz$

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

$=ka^3$

$\int_{S_3} \vec{B}.d \vec{A}=0$

$\int_{S_4} \vec{B}.d \vec{A}=0$

$\int_{S_5} \vec{B}.d \vec{A}=0$

$\int_{S_6} \vec{B}.d \vec{A}=0$

$\oint \vec{B}.d \vec{A}=2ka^3 \neq 0$

Hence $\vec{F}$ cannot repeat a negative fields.

Calculate the magnetic field produced by a finite current element as shown

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

$\vec{dl} \times \vec{s}=dl s \sin \theta$

$\sin \theta = \cos \alpha$

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

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

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

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

$|\vec{B}|=\frac{\mu_0 I}{4 \pi} \int_{z_1}^{z-2} \frac{dz \cos \alpha}{(r^2=z^2)}$

$z=r \tan \alpha$

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

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

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

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

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

$|\vec{B}|=[\frac{\mu_0I}{4 \pi r}[\frac{z^2}{\sqrt{z_2^2 + r^2}}-\frac{z_1}{\sqrt{z_1^2+r^2}}]$

$z_1 = -L/8$

$z_2 =+L/8$

$r=L/8$

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

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

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 $=4 \times \frac{2 \mu_0 I}{\pi L}\sqrt{2}$

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

$Circle$

$L=2 \pi R$

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

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

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

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

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

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

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

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

$=\frac{8 \sqrt{2}}{\pi^2}$

$\simeq 1.15$

Component of magnetic field along the verticle direction

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

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

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

Component of magnetic field along the horizontal direction

$B_m= \frac{\mu_0I}{2a}[0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1\frac{\sqrt{3}}{2}+\frac{1}{2}]$

$\frac{\mu_0I}{2a}[2+\sqrt{3}]$

$|\vec{B}|=[B_v^2+B_H^2]^{1/2}$

$=\frac{\mu_0I}{2a}[1+(2+\sqrt{2})^2]^{1/2}$

$\simeq 1.93 \frac{\mu_0 I}{a}$

Current density

$I=\frac{I}{\pi(R^2-a^2)}$

Magnetic field due to solid conduction (without the hole) of radius R

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

$2 \pi r B = \mu_0 J. \pi r^2$

$B=\frac{\mu_0 Jr}{2}; \vec{B}=\frac{\mu_0 Jr}{2}\hat{m}$

Magnetic field due to solid cyclinder of radius a at a distance from the centre

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

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

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

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

Total field

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

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

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

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

$\vec{B}=\frac{\mu_0 J b}{2}\hat{j}$

$=\frac{\mu_0 I b}{2 \pi (R^2 - a^2)}\hat{j}$

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

$\lambda = 0.5 \mu m$

$\vec{B}=\frac{(15 \hat{i} - 10 \hat{j})}{c} \sin [4 \pi \times 10^6 (ct-2)]T$