Two long parallel conductors S_{1} and S_{2} | vedclass

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(C) The magnetic field $B_{1}$ due to conductor $S_{1}$ at point $P$ (distance $r_{1} = 4 \, cm = 0.04 \, m$) is directed into the page ($-\hat{k}$ di

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$3$

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(C) The magnetic field $B_{1}$ due to conductor $S_{1}$ at point $P$ (distance $r_{1} = 4 \, cm = 0.04 \, m$) is directed into the page ($-\hat{k}$ direction):$B_{1} = \frac{\mu_{0} I_{1}}{2 \pi r_{1}} = \frac{\mu_{0} 	imes 4}{2 \pi 	imes 0.04} = \frac{\mu_{0}}{2 \pi} 	imes 100 \, T$ (in $-\hat{k}$ direction).The magnetic field $B_{2}$ due to conductor $S_{2}$ at point $P$ (distance $r_{2} = 10 \, cm - 4 \, cm = 6 \, cm = 0.06 \, m$) is directed out of the page ($+\hat{k}$ direction):$B_{2} = \frac{\mu_{0} I_{2}}{2 \pi r_{2}} = \frac{\mu_{0} 	imes 2}{2 \pi 	imes 0.06} = \frac{\mu_{0}}{2 \pi} 	imes \frac{100}{3} \, T$ (in $+\hat{k}$ direction).The net magnetic field $\overrightarrow{B}_{net} = B_{1} + B_{2} = \frac{\mu_{0}}{2 \pi} \left( -100 + \frac{100}{3} ight) \hat{k} = \frac{\mu_{0}}{2 \pi} \left( -\frac{200}{3} ight) \hat{k} = -\frac{100 \mu_{0}}{3 \pi} \hat{k} \, T$.The Lorentz force is $\overrightarrow{F} = q(\overrightarrow{v} 	imes \overrightarrow{B}) = 3 \pi \left[ (2 \hat{i} + 3 \hat{j}) 	imes \left( -\frac{100 \mu_{0}}{3 \pi} \hat{k} ight) ight]$.Using $\mu_{0} = 4 \pi 	imes 10^{-7} \, T \cdot m/A$,we have $\frac{\mu_{0}}{2 \pi} = 2 	imes 10^{-7}$.$\overrightarrow{F} = 3 \pi 	imes \left( -\frac{200}{3} 	imes 10^{-7} ight) [ 2(\hat{i} 	imes \hat{k}) + 3(\hat{j} 	imes \hat{k}) ]$.Since $\hat{i} 	imes \hat{k} = -\hat{j}$ and $\hat{j} 	imes \hat{k} = \hat{i}$:$\overrightarrow{F} = -200 \pi 	imes 10^{-7} [ -2 \hat{j} + 3 \hat{i} ] = 2 \pi 	imes 10^{-5} [ 2 \hat{j} - 3 \hat{i} ] = 4 \pi 	imes 10^{-5} [ -1.5 \hat{i} + \hat{j} ]$.Comparing this with $4 \pi 	imes 10^{-5} (-x \hat{i} + 2 \hat{j})$,we find $x = 3$.

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