Question

A current $i=2.5\text{A}$ flows along the circular coil of three turns in anticlockwise direction. The equation of circle is given by $x^2+y^2=9{\text{cm}}^2$ $(x$ & $y$ are in $\text{cm})$. The magnetic field at point $P(0,0,4\text{cm})$ is :
[consider $x$ and $y$ axis in the plane of paper and perpendicular outward direction as $+z$ - axis ]

• A. $(36\pi \times {10}^{-7}\text{T})\hat{k}$
• B. $(108\pi \times {10}^{-7}\text{T})\hat{k}$
• C. $(\frac{9\pi }{5}\times {10}^{-7}\text{T})(-\hat{k})$
• D. $(36\pi \times {10}^{-7}\text{T})(-\hat{k})$

Answer 1

The correct option is B $(108\pi \times {10}^{-7}\text{T})\hat{k}$

From the equation of circle,
$x^2+y^2=9{\text{cm}}^2$
Comparing it with $x^2+y^2=a^2$

We find that the circular coil is placed in $x-y$ plane with centre at origin $(0,0)$ and radius $R=3\text{cm}$.

From the shown figure it is clear that point $\text{P}(x=0,y=0,z=4\text{cm})$ lies along the axis of coil line (i.e. $z-$axis)

Using the relation;

$B_P=\frac{\left({\mu }_{0}i{R}^2\right)n}{2(R^2+x^2{)}^{3/2}}$

$B_P=\frac{(4\pi \times {10}^{-7})\times 2.5\times (3\times {10}^{-2}{)}^2\times 3}{2\left[\right(3\times {10}^{-2}{)}^2+(4\times {10}^{-2}{)}^2{]}^{3/2}}$

$B_P=\frac{27\times 10\pi \times {10}^{-11}}{2\left[\right(25\times {10}^{-4}){]}^{3/2}}$

$B_P=108\pi \times {10}^{-7}\text{T}$

Also, from right-hand thumb rule direction of field at $\text{P}$ is along $+z$- axis.

$\therefore \overrightarrow{B_P}=(108\pi \times {10}^{-7})\hat{k}$

Ans : $\left(b\right)$

Why this question ? Tip: Just focus on the equation of circle, it will give the clue for plane of coil $\left(XY\right)$, centre $(0,0)$ & the radius $\left(R\right)$. Thus, we can easily visualize the point $\text{P}$ and apply the formula for magnetic field due to coil.