Two straight long wires fixed in the X-Y plane along the lines x = -a and x = a carry steady current of magnitude I in the +Y and -Y directions respectively. Let the magnetic field at a point p(0,0,z) on the positive Z-axis, and at a point Q(0,0,-z) on the negative Z-axis be ${\vec{B}}_{P} a n d {\vec{B}}_{Q}$ respectively. Which of the following statements is/are correct?

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${\vec{B}}_{P} a n d {\vec{B}}_{Q}$ are respectively along the –Z and +Z directions

${\vec{B}}_{P} a n d {\vec{B}}_{Q}$ are both along the –Z direction

$| {\vec{B}}_{P} | a n d | {\vec{B}}_{Q} |$ are both proportional to $\frac{1}{z^{2}} if z=a$

$| {\vec{B}}_{P} | a n d | {\vec{B}}_{Q} |$ are both proportional to $\frac{1}{z^{3}} if z=a$

answer is B.

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Detailed Solution

Remember that the +Z direction is along the direction $(+ X) \times (+ Y)$ i.e. $\hat{k} = \hat{i} \times \hat{j}$

We can visualise this by drawing a diagram showing the X-Z plane such that the wires appear points. The field due to the wires will only have X and Z components, and its Y-component is zero. At any point on the Z-axis (the positive Z-axis or the negative Z-axis), the X-components of the field due to each wire cancel and the –Z components add up
At the point $P (0, 0, z), B_{P} = 2 (\frac{μ_{0} I}{2 π \sqrt{a^{2} + z^{2}}}) (\frac{a}{\sqrt{a^{2} + z^{2}}}) = \frac{μ_{0} I a}{2 (a^{2} + z^{2})}$