Answer

Verified

394.2k+ views

**Hint**To solve this question, we have to use the formula for the magnetic field due to an infinitely long carrying wire. And for getting the direction, we need to use the right hand thumb rule.

The formula which is used in solving this question is given by

$\Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}}$ , here $B$ is the magnetic field produced by an infinitely long wire carrying a current $I$ at a perpendicular distance of $r$ from it.

**Complete step by step answer**

Let us consider a point P on the x-axis at a distance of $x$ from the origin, as shown in the following figure.

We know that the magnetic field due to an infinitely long wire is given by

$\Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi r}}$ (1)

So we need to calculate the perpendicular distance of the point P from each of the wires A, B and C.

For wire A:

Consider the following figure.

From the figure, we see that the perpendicular distance of the point P is

$\Rightarrow {r_1} = PM$

In the triangle $OPM$, we have

$\Rightarrow PM = OP\cos {45^ \circ }$

$\Rightarrow {r_1} = x\cos {45^ \circ }$

So we get

$\Rightarrow {r_1} = \dfrac{x}{{\sqrt 2 }}$ (2)

So from (1) the magnetic field due to wire A at the point P is

$\Rightarrow {B_A} = \dfrac{{{\mu _0}I}}{{2\pi {r_1}}}$

From (2)

$\Rightarrow {B_A} = \dfrac{{{\mu _0}I}}{{2\pi \dfrac{x}{{\sqrt 2 }}}}$

$\Rightarrow {B_A} = \dfrac{{{\mu _0}I}}{{\sqrt 2 \pi x}}$

By the right hand thumb rule, the direction of this field is along the positive y-axis. So in terms of the unit vector we have

$\Rightarrow {B_A} = \dfrac{{{\mu _0}I}}{{\sqrt 2 \pi x}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{y} $

Similarly, the magnetic field due to the wire B at the point P

$\Rightarrow {B_B} = \dfrac{{{\mu _0}I}}{{\sqrt 2 \pi x}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{y} $

Also, the magnetic field at P due to wire C is

$\Rightarrow {B_C} = \dfrac{{{\mu _0}I}}{{2\pi x}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{y} $

So the net magnetic field at point P is

$\Rightarrow B = {B_A} + {B_B} + {B_C}$

From (3) (4) and (5)

$\Rightarrow B = \dfrac{{{\mu _0}I}}{{\sqrt 2 \pi x}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{y} + \dfrac{{{\mu _0}I}}{{\sqrt 2 \pi x}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{y} + \dfrac{{{\mu _0}I}}{{2\pi x}}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{y} $

On simplifying we get

$\Rightarrow B = \dfrac{{{\mu _0}I}}{{2\pi x}}\left( {1 + 2\sqrt 2 } \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{y} $

Thus, this is the required magnetic field as a function of $x$.

**Hence, the correct answer is option (C).**

**Note**

The given set of axes is not the standard set of coordinate axes. So, for getting the direction of the positive y-axis, we use the right hand thumb rule. We should not assume the direction by ourselves.

Recently Updated Pages

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

The provincial president of the constituent assembly class 11 social science CBSE

Gersoppa waterfall is located in AGuyana BUganda C class 9 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

The hundru falls is in A Chota Nagpur Plateau B Calcutta class 8 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE