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Question

(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Let the wire carrying $I_1 = 8 \ A$

Vedclass · Accepted Answer

$8 	imes 10^{-5} \ T$

Vedclass · Answer

(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Let the wire carrying $I_1 = 8 \ A$ be at $x = 0$ and the wire carrying $I_2 = 10 \ A$ be at $x = 9 \ cm$.The point of interest is at $x = 4 \ cm$.For the first wire $(I_1 = 8 \ A)$: $r_1 = 4 \ cm = 0.04 \ m$. The magnetic field $B_1 = \frac{\mu_0 	imes 8}{2 \pi 	imes 0.04} = \frac{2 	imes 10^{-7} 	imes 8}{0.04} = 4 	imes 10^{-5} \ T$.For the second wire $(I_2 = 10 \ A)$: $r_2 = 9 \ cm - 4 \ cm = 5 \ cm = 0.05 \ m$. The magnetic field $B_2 = \frac{\mu_0 	imes 10}{2 \pi 	imes 0.05} = \frac{2 	imes 10^{-7} 	imes 10}{0.05} = 4 	imes 10^{-5} \ T$.Since the currents are in opposite directions, by the right-hand rule, the magnetic fields at the point between the wires will be in the same direction.Therefore, the net magnetic field $B_{net} = B_1 + B_2 = 4 	imes 10^{-5} \ T + 4 	imes 10^{-5} \ T = 8 	imes 10^{-5} \ T$.

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