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# Two concentric coils each of radius equal to $2\pi cm$ are placed at right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively, the magnetic induction (in $Wb{{m}^{-2}}$) at the centre of the coils will beA. $12\times {{10}^{-5}}$ B. ${{10}^{-5}}$ C. $5\times {{10}^{-5}}$ D. $7\times {{10}^{-5}}$

Last updated date: 13th Jun 2024
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Hint: We are given two concentric coils placed at right angles with each other which implies that the axis of both the coils and thus their respective magnetic fields must also be perpendicular to each other. Therefore, we shall find the magnetic fields due to both the coils individually and then find their resultant net value of magnetic induction at their centre.

Complete step by step solution:
The magnetic field, $B$ due to a circular coil is given as:
$B=\dfrac{{{\mu }_{0}}i}{2r}$
Where,
${{\mu }_{0}}=$ magnetic permeability constant and ${{\mu }_{0}}=4\pi \times {{10}^{-7}}\,Wb{{A}^{-1}}{{m}^{-1}}$
$i=$ current in the coil
$r=$ radius of coil

For coil 1, we have $i=3A$ and $r=2\pi cm$ or $r=2\pi \times {{10}^{-2}}m$,
Applying the above formula of magnetic field, we get
${{B}_{1}}=\dfrac{{{\mu }_{0}}\left( 3 \right)}{2\left( 2\pi \times {{10}^{-2}} \right)}$
Now, we shall put the value the magnetic permeability constant and then we get,
$\Rightarrow {{B}_{1}}=\dfrac{\left( 4\pi \times {{10}^{-7}} \right)\left( 3 \right)}{2\left( 2\pi \times {{10}^{-2}} \right)}$
Here, we cancel $4\pi$ and ${{10}^{-2}}$ from the numerator and denominator.
$\Rightarrow {{B}_{1}}=3\times {{10}^{-5}}$ ……………….. (1)
Similarly, for coil 2, we have $i=4A$ and $r=2\pi cm$ or $r=2\pi \times {{10}^{-2}}m$,
Applying the above formula of magnetic field, we get
${{B}_{2}}=\dfrac{{{\mu }_{0}}\left( 4 \right)}{2\left( 2\pi \times {{10}^{-2}} \right)}$
Now, we shall put the value the magnetic permeability constant and then we get,
$\Rightarrow {{B}_{2}}=\dfrac{\left( 4\pi \times {{10}^{-7}} \right)\left( 4 \right)}{2\left( 2\pi \times {{10}^{-2}} \right)}$
Here, we cancel $4\pi$ and ${{10}^{-2}}$ from the numerator and denominator.
$\Rightarrow {{B}_{2}}=4\times {{10}^{-5}}$ ………………. (2)
Now, we find the resultant net magnetic field using the formula $B=\sqrt{B_{1}^{2}+B_{2}^{2}}$.
Substituting the values of ${{B}_{1}}$ and ${{B}_{2}}$ from (1) and (2), we get
$\Rightarrow B=\sqrt{{{\left( 3\times {{10}^{-5}} \right)}^{2}}+{{\left( 4\times {{10}^{-5}} \right)}^{2}}}$
Taking ${{\left( {{10}^{-5}} \right)}^{2}}$ common and finding their square, we get
\begin{align} & \Rightarrow B=\sqrt{{{\left( {{10}^{-5}} \right)}^{2}}\left( {{3}^{2}}+{{4}^{2}} \right)} \\ & \Rightarrow B={{10}^{-5}}\sqrt{{{3}^{2}}+{{4}^{2}}} \\ \end{align}
We know that ${{3}^{2}}=3\times 3=9$ and ${{4}^{2}}=4\times 4=16$. Thus, substituting these values and solving further.
\begin{align} & \Rightarrow B={{10}^{-5}}\sqrt{9+16} \\ & \Rightarrow B={{10}^{-5}}\sqrt{25} \\ & \Rightarrow B=5\times {{10}^{-5}} \\ \end{align}
Therefore, the magnetic induction at centre is $5\times {{10}^{-5}}Wb{{m}^{-2}}$.

Hence, the correct option is (C) $5\times {{10}^{-5}}$.

Note:
The magnetic field is a vector quantity which has both magnitude and direction. Thus, the resultant of two or more magnetic field values is calculated by using rules of vector addition. Since, most of the quantities that we deal with in physics are vectors, therefore, one must have appropriate knowledge of vectors beforehand only.