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Physics Question on Moving charges and magnetism

A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic vector field B at a point having coordinate (x, y ) on the z = 0 plane is

A

μ0I(yi^−xj^)2π(x2+y2)\frac{\mu_0 I(y\widehat{i}-x\widehat{j})}{2\pi(x^2+y^2)}

B

μ0I(yi^+xj^)2π(x2+y2)\frac{\mu_0 I(y\widehat{i}+x\widehat{j})}{2\pi(x^2+y^2)}

C

μ0I(xj^−xi^)2π(x2+y2)\frac{\mu_0 I(x\widehat{j}-x\widehat{i})}{2\pi(x^2+y^2)}

D

μ0I(xi^−yj^)2π(x2+y2)\frac{\mu_0 I(x \widehat{i}-y\widehat{j})}{2\pi(x^2+y^2)}

Answer

μ0I(yi^−xj^)2π(x2+y2)\frac{\mu_0 I(y\widehat{i}-x\widehat{j})}{2\pi(x^2+y^2)}

Explanation

Solution

Magnetic field at P is B, perpendicular to OP in the direction
shown in figure.
So,\hspace15mm B=B\, sin\, \theta \widehat{i}-Bcos\, \theta\widehat{j}
Here,          B=μ0I2πr\, \, \, \, \, \, \, \, \, \, B=\frac{\mu_0 I}{2\pi r}
                 sinθ=yr       and         cosθ=xr\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, sin\theta=\frac{y}{r}\, \, \, \, \, \, \, and\, \, \, \, \, \, \, \, \, cos\theta=\frac{x}{r}
∴             B=μ0I2π.1r2(yi^−xi^)\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, B=\frac{\mu_0 I}{2\pi}.\frac{1}{r^2}(y\widehat{i}-x\widehat{i})
                 =μ0I(yi^−xj^)2π(x2+y2\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{\mu_0 I(y\widehat{i}-x\widehat{j})}{2\pi(x^2+y^2}
                                                                (asx2+y2)\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (as x^2+y^2)