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(B) The magnetic field $B$ due to an infinitely long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For the wire carrying $9

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$\frac{5\mu_0}{2\pi} \otimes$

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(B) The magnetic field $B$ due to an infinitely long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2\pi r}$.For the wire carrying $9 	ext{ A}$ current (upward),the point $M$ is at a distance $r_1 = 2 	ext{ m}$ to the right. Using the right-hand thumb rule,the direction of the magnetic field at $M$ is into the page $(\otimes)$.$B_1 = \frac{\mu_0 (9)}{2\pi (2)} = \frac{9\mu_0}{4\pi} \otimes$.For the wire carrying $3 	ext{ A}$ current (downward),the point $M$ is at a distance $r_2 = 4 + 2 = 6 	ext{ m}$ to the right. Using the right-hand thumb rule,the direction of the magnetic field at $M$ is into the page $(\otimes)$.$B_2 = \frac{\mu_0 (3)}{2\pi (6)} = \frac{3\mu_0}{12\pi} = \frac{\mu_0}{4\pi} \otimes$.Since both magnetic fields are in the same direction (into the page),the net magnetic field $B_{net}$ is:$B_{net} = B_1 + B_2 = \frac{9\mu_0}{4\pi} + \frac{\mu_0}{4\pi} = \frac{10\mu_0}{4\pi} = \frac{5\mu_0}{2\pi} \otimes$.

Calculate the magnetic field at point $M$ for the given current distribution.

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