Magnetic Field of a Straight Wire

# Magnetic field of a straight wire: → of any given length. \[d\vec{B}=\frac{\mu_{0}I}{4\pi}(\frac{\vec{dl}\times\hat{r}}{\gamma^{2}})\] X P \[\vec{r}\] • \(\vec{dl}\times\hat{r}=dl~\sin\alpha~\hat{\theta}\) BOI gre : \(180^{\circ}-\alpha+90+\theta=180^{\circ}\) \[\alpha=90+\theta\] \[dl=s\sec^{2}\theta d\theta\] \[\gamma=\sqrt{s^{2}+l^{2}}\] \[\frac{1}{s}=\tan~\theta\] \[\Rightarrow\vec{dB}=\frac{\mu_{0}I}{4\pi}\frac{s\sec^{2}\theta d\theta~\sin(90+\theta)}{s^{2}(1+\tan^{2}\theta)}\] I I D \(\vec{B}=\frac{\mu_{0}I}{4\pi s}[sin~\theta_{2}+sin~\theta_{1}](\) Type \[\vec{B}=\frac{\mu_{0}I}{4\pi s}\int_{-\theta_{1}}^{\theta_{2}}cos~\theta~d\theta~\theta~\theta~d~\theta~\theta\\ +\theta_{1}\sqrt{\vec{B}=\frac{\mu}{4\pi}}\] \[\vec{dB}=\frac{\mu_{0}I}{4\pi\epsilon}cos\theta d\theta\theta\] \[B^{3}=\frac{u_{0}I}{4\pi\lambda}(sin~\theta_{2}-sin~\theta_{1}))\] \[d\vec{B}=\frac{\mu_{0}I}{4\pi}\frac{Ssec^{2}\theta d\theta~cos\theta~\theta}{8^{2}(1+tan^{2}\theta)}\] \[\begin{cases}\cdot\cdot\cdot\frac{J^{n}}{S^{2}}=tan^{2}\theta\] \[\theta_{2}\] A 01 82 OB OA B = \[sin(-\theta)\] \[=-sin~\theta\] \[\vec{B}=\frac{\omega_{0}I}{4\pi S}[sin~\theta_{L}+sin~\theta_{1}]\] Find mag-field, at contre of D \[\vec{B}_{0}=?\] a MoI [Sin45°+ Sim45] \[\vec{B}=\frac{\mu_{0}I}{4\pi S}[sin~\theta_{2}+sin~\theta_{i}]\] eg. 21-33 Pe 43 A B \[\vec{B}_{AB}=\frac{\mu_{0}I}{\pi\sqrt{2}a}C\] for last eg. (page-115) \[\vec{B}_{AB}=\frac{\mu_{0}I}{4\pi\beta}[sin\theta_{2}+sin\theta_{1}]\] \[=\frac{\mu_{0}I}{4\pi\beta}[1+\frac{1}{\sqrt{2}}]\otimes\] \[\vec{B}_{PP}=\frac{\mu_{0}I}{4\pi\beta}[sin\theta_{2}-sin\theta_{1}]\] \[=0\] # ∴ \(\vec{B}=\frac{\mu_{0}I}{4\pi\beta}[sin\theta_{2}+sin\theta_{1}]\) Limiting Case:- ① Infinite wire: \[\vec{B}_{infinite}=\frac{\mu_{0}I}{2\pi\beta}\otimes\] ② Semi-infinite wire→ \[\vec{B}_{P_{0}}=\frac{\mu_{0}I}{4\pi\beta}\otimes\]