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(B) The magnetic field due to an infinitely long straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For the left w

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$1 	imes 10^{-4} \ T$

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(B) The magnetic field due to an infinitely long straight wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For the left wire, the point $O$ lies on the axis of the horizontal segment, so the magnetic field due to the horizontal segment at $O$ is $0$. The vertical segment is a semi-infinite wire at a distance $r = 2 \ cm = 0.02 \ m$. The magnetic field due to a semi-infinite wire is $B_1 = \frac{\mu_0 I}{4\pi r}$.Using the right-hand rule, the field $B_1$ at $O$ points into the page $(\otimes)$.Similarly, for the right wire, the point $O$ lies on the axis of the horizontal segment, so its contribution is $0$. The vertical segment is a semi-infinite wire at a distance $r = 2 \ cm = 0.02 \ m$. The magnetic field $B_2$ at $O$ points into the page $(\otimes)$.Total magnetic field $B = B_1 + B_2 = 2 	imes \left( \frac{\mu_0 I}{4\pi r} ight) = \frac{\mu_0 I}{2\pi r}$.Substituting the values: $B = \frac{(4\pi 	imes 10^{-7}) 	imes 10}{2\pi 	imes 0.02} = \frac{2 	imes 10^{-6}}{0.02} = 10^{-4} \ T$.

Two infinitely long wires, each carrying a current $I = 10 \ A$, are bent to form a right angle as shown in the figure. Find the magnetic induction at point $O$. $\left[\mu_0 = 4\pi \times 10^{-7} \ H \ m^{-1}\right]$

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