Question

Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart. The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the midpoint is –
(a) $\frac{\mu_{0}id}{\pi(d^{2}+x^{2})}$
(b) $\frac{\mu_{0}ix}{\pi(d^{2}-x^{2})}$
(c) $\frac{\mu_{0}ix}{(d^{2}+x^{2})}$
(d) $\frac{\mu_{0}id}{(d^{2}+x^{2})}$

Answer 1

Given:

Two long parallel wires carry equal current i flowing in the same direction are at a distance 2d apart.

To find:

The magnetic field B at a point lying on the perpendicular line joining the wires and at a distance x from the midpoint.

Calculation:

Net Magnetic Field at Point P :

$\sf{ \therefore \: B_{net} = \dfrac{ \mu_{0}}{4\pi} \bigg \{\dfrac{2i}{(d - x)} \bigg \} - \dfrac{ \mu_{0}}{4\pi} \bigg \{\dfrac{2i}{(d + x)} \bigg \}}$

$\sf{ = > \: B_{net} = \dfrac{ \mu_{0}}{4\pi} \bigg \{\dfrac{2i}{(d - x)} - \dfrac{2i}{(d + x)} \bigg \}}$

$\sf{ = > \: B_{net} = \dfrac{ \mu_{0} \times 2i}{4\pi} \bigg \{\dfrac{1}{(d - x)} - \dfrac{1}{(d + x)} \bigg \}}$

$\sf{ = > \: B_{net} = \dfrac{ \mu_{0}i}{2\pi} \bigg \{\dfrac{1}{(d - x)} - \dfrac{1}{(d + x)} \bigg \}}$

$\sf{ = > \: B_{net} = \dfrac{ \mu_{0}i}{2\pi} \bigg \{\dfrac{(d + x) - (d - x)}{ {d}^{2} - {x}^{2} } \bigg \}}$

$\sf{ = > \: B_{net} = \dfrac{ \mu_{0}i}{2\pi} \bigg \{\dfrac{2x}{ {d}^{2} - {x}^{2} } \bigg \}}$

$\sf{ = > \: B_{net} = \dfrac{ \mu_{0}i}{\pi} \bigg \{\dfrac{x}{ {d}^{2} - {x}^{2} } \bigg \}}$

So , final answer is :

$\boxed{ \bold{ \red{ \: B_{net} = \dfrac{ \mu_{0}i}{\pi} \bigg \{\dfrac{x}{ {d}^{2} - {x}^{2} } \bigg \}}}}$