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For adjoining fig., the magnetic field at point 'P' will be :
A
$\dfrac{\mu_0}{4\pi}$
B
$\dfrac{\mu_0}{\pi}\otimes$
C
$\dfrac{\mu_0}{2\pi}\otimes$
D
$\dfrac{\mu_0}{2\pi}$
Detailed Solution
Magnetic field of a long straight wire: $B = \dfrac{\mu_0I}{2\pi r}$
Due to the 5 A wire at 2.5 m: $B_1 = \dfrac{5\mu_0}{2\pi \times 2.5} = \dfrac{2\mu_0}{2\pi}$, directed into the page ($\otimes$) at P.
Due to the 2.5 A wire at 2.5 m: $B_2 = \dfrac{2.5\mu_0}{2\pi \times 2.5} = \dfrac{\mu_0}{2\pi}$, directed out of the page at P (opposite, since P is on the other side of this wire).
Resultant: $B = \dfrac{2\mu_0}{2\pi} - \dfrac{\mu_0}{2\pi} = \dfrac{\mu_0}{2\pi}$, into the page ($\otimes$).
Due to the 5 A wire at 2.5 m: $B_1 = \dfrac{5\mu_0}{2\pi \times 2.5} = \dfrac{2\mu_0}{2\pi}$, directed into the page ($\otimes$) at P.
Due to the 2.5 A wire at 2.5 m: $B_2 = \dfrac{2.5\mu_0}{2\pi \times 2.5} = \dfrac{\mu_0}{2\pi}$, directed out of the page at P (opposite, since P is on the other side of this wire).
Resultant: $B = \dfrac{2\mu_0}{2\pi} - \dfrac{\mu_0}{2\pi} = \dfrac{\mu_0}{2\pi}$, into the page ($\otimes$).