Two long straight wires are parallel and carry current in the same direction.The currents are 8.0 and 12 A and the wires are separated by 0.40 cm.The magnetic field at a point midway between the wires is: A)0 T B)4.0 *10^-4 T C)8.0 * 10^-4 T D)12 * 10^-4 T E)20 * 10^-4 T

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The magnetic field due to a long straight wire carrying current is given by:$$B = \frac{\mu_0 I}{2\pi r}$$where:* B is the magnetic field in Tesla* μ0 is the permeability of free space (4π × 10^-7 T·m/A)* I is the current in Amperes* r is the distance from the wire in metersIn this problem, we have two wires carrying currents of 8.0 A and 12 A, respectively. The wires are separated by 0.40 cm, or 0.0040 m. The magnetic field at a point midway between the wires is the sum of the magnetic fields due to each wire.$$B = B_1 + B_2$$$$B = \frac{\mu_0 I_1}{2\pi r_1} + \frac{\mu_0 I_2}{2\pi r_2}$$$$B = \frac{(4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A})(8.0 	ext{ A})}{2\pi (0.0020 	ext{ m})} + \frac{(4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A})(12 	ext{ A})}{2\pi (0.0020 	ext{ m})}$$$$B = 4.0 	imes 10^{-4} 	ext{ T}$$Therefore, the correct answer is B) 4.0 × 10^-4 T.

Two Long Straight Wires Are Parallel and Carry Current in the Same