Two long straight parallel conductors A and B | vedclass

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(C) Given: Current in conductor $A$,$I_1 = 4.5 \ A$. Current in conductor $B$,$I_2 = 8 \ A$. Distance of point $P$ from $A$,$r_1 = 15 \ cm = 0.15 \ m$

Vedclass · Accepted Answer

$10^{-5} \ T$

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(C) Given: Current in conductor $A$,$I_1 = 4.5 \ A$. Current in conductor $B$,$I_2 = 8 \ A$. Distance of point $P$ from $A$,$r_1 = 15 \ cm = 0.15 \ m$. Distance of point $P$ from $B$,$r_2 = 10 \ cm = 0.10 \ m$.The magnetic field due to a long straight wire is given by $B = \frac{\mu_0 I}{2 \pi r}$.Using the right-hand rule,the magnetic field $B_1$ due to conductor $A$ at point $P$ is directed into the page,and the magnetic field $B_2$ due to conductor $B$ at point $P$ is directed out of the page.$B_1 = \frac{\mu_0 I_1}{2 \pi r_1} = \frac{2 	imes 10^{-7} 	imes 4.5}{0.15} = 6 	imes 10^{-6} \ T$ (into the page).$B_2 = \frac{\mu_0 I_2}{2 \pi r_2} = \frac{2 	imes 10^{-7} 	imes 8}{0.10} = 16 	imes 10^{-6} \ T$ (out of the page).The resultant magnetic field $B = B_2 - B_1 = (16 - 6) 	imes 10^{-6} \ T = 10 	imes 10^{-6} \ T = 10^{-5} \ T$.

Two long straight parallel conductors $A$ and $B$ carrying currents $4.5 \ A$ and $8 \ A$ respectively are separated by $25 \ cm$ in air. The resultant magnetic field at a point $P$ which is at a distance of $15 \ cm$ from conductor $A$ and $10 \ cm$ from conductor $B$ is:

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