There are two infinitely long straight current carrying conductors and they are held at right angle to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is

Option 1 -

$\frac{μ_{0} Ι}{4 π x y} [\sqrt[]{x^{2} + y^{2}} - (x + y)]$

Option 2 -

$\frac{μ_{0} Ι}{4 π x y} [\sqrt[]{x^{2} + y^{2}} + (x + y)]$

Option 3 -

$\frac{μ_{0} Ι x y}{4 π} [\sqrt[]{x^{2} + y^{2}} - (x + y)]$

Option 4 -

$\frac{μ_{0} Ι x y}{4 π} [\sqrt[]{x^{2} + y^{2}} + (x + y)]$

2 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$B_{1} = \frac{μ_{0} l}{4 π y} [s i n θ_{1} + s i n 90 °] - - - - - - - (i)$

B due to OX wire

$B_{2} = \frac{μ_{0} l}{4 π x} [s i n θ_{2} + s i n 90 °]$

As in the diagram direction of B₁ and B₂ are in downward direction

So B = B₁ + B₂

$B = \frac{μ_{0} l}{4 π y} (s i n θ_{1} + 1) + \frac{μ_{0} l}{4 π x} (s i n θ_{2} + 1)$

$B = \frac{μ_{0} l}{4 π x y} [(x + y) + \sqrt[]{x^{2} + y^{2}}]$

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There are two infinitely long straight current carrying conductors and they are held at right angle to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is

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