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(B) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.At the mid-point $O$,the distance f

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$30 \, A$ in the opposite direction

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(B) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.At the mid-point $O$,the distance from each wire is $r = 4 \, cm = 4 	imes 10^{-2} \, m$.Since the total magnetic field is $300 \, \mu T = 3 	imes 10^{-4} \, T$,and the currents are in opposite directions (as shown in the figure),the magnetic fields due to both wires at the mid-point point in the same direction.Thus,$B_{total} = B_1 + B_2 = 2 	imes \frac{\mu_0 I}{2 \pi r}$.Substituting the values: $3 	imes 10^{-4} = 2 	imes \frac{4 \pi 	imes 10^{-7} 	imes I}{2 \pi 	imes 4 	imes 10^{-2}}$.$3 	imes 10^{-4} = \frac{2 	imes 10^{-7} 	imes I}{2 	imes 10^{-2}} = 10^{-5} 	imes I$.$I = \frac{3 	imes 10^{-4}}{10^{-5}} = 30 \, A$.Since the fields add up to a non-zero value at the mid-point,the currents must be in opposite directions.

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