Question

Two coils ‘P’ and ‘Q’ are separated by some distance. When a current of 3 A flows through coil ‘P’, a magnetic flux \[{{10}^{-3}}\,Wb\] passes through ‘Q’. No current is passed through ‘Q’. When no current passes through ‘P’ and a current of 2A passes through ‘Q’, the flux through ‘P’ is?

\[A.\,6.67\times {{10}^{-3}}\,Wb\]
\[B.\,6.67\times {{10}^{-4}}\,Wb\]
\[C.\,3.67\times {{10}^{-4}}\,Wb\]
\[D.\,3.67\times {{10}^{-3}}\,Wb\]

Answer 1

Hint: The formula that relates the electric flux through a coil, the magnetic field produced and the area of the coil should be used to solve this problem. The equations of the flux of the coils P and Q should be divided and the given values of the currents and the flux through the coil Q should be substituted to find the value of the flux through the coil P.
Formula used:
\[\begin{align}
  & \phi =BA \\
 & \Rightarrow \phi =\dfrac{{{\mu }_{0}}i{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}\times \pi {{r}^{2}} \\
\end{align}\]

Complete answer:
The formula for calculating the magnetic flux is given by the formula as follows.
\[\phi =BA\]
Where B is the magnetic field produced and A is the area of the coil.
The magnetic field produced and the area of the coil equation can be expressed as,
\[\phi =\dfrac{{{\mu }_{0}}i{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}\times \pi {{r}^{2}}\]
Where R is the radius of the one coil, i is the current through the coil and r is the radius of the other coil.
Represent the above equations in terms of the coils P and Q. So, we get,
\[\begin{align}
  & {{\phi }_{q}}=\dfrac{{{\mu }_{0}}i{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}\times \pi {{r}^{2}} \\
 & {{\phi }_{p}}=\dfrac{{{\mu }_{0}}i{{r}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}\times \pi {{R}^{2}} \\
\end{align}\]
From the data, we have the data as follows.
When a current of 3 A flows through coil ‘P’, a magnetic flux \[{{10}^{-3}}\,Wb\] passes through ‘Q’. No current is passed through ‘Q’. When no current passes through ‘P’ and a current of 2A passes through ‘Q’
So, the equations are as follows.
\[\begin{align}
  & {{\phi }_{q}}=\dfrac{{{\mu }_{0}}(2){{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}\times \pi {{r}^{2}}={{10}^{-3}} \\
 & {{\phi }_{p}}=\dfrac{{{\mu }_{0}}(3){{r}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{{}^{3}/{}_{2}}}}\times \pi {{R}^{2}} \\
\end{align}\]
Divide the above equations in order to simplify this complicated equation.
\[\begin{align}
  & \dfrac{{{\phi }_{p}}}{{{\phi }_{q}}}=\dfrac{2}{3}\dfrac{{{({{R}^{2}}+{{x}^{2}})}^{\dfrac{3}{2}}}}{{{({{r}^{2}}+{{x}^{2}})}^{\dfrac{3}{2}}}} \\
 & \Rightarrow \dfrac{{{\phi }_{p}}}{{{10}^{-3}}}=\dfrac{2}{3} \\
\end{align}\]
Continue the further calculation to find the value of the flux through the coil P.
\[{{\phi }_{p}}=6.67\times {{10}^{-4}}\,Wb\]
Therefore, the value of the flux through the coil P when the current of 2 A passes through the coil Q is \[6.67\times {{10}^{-4}}\,Wb\]
As, the value of the flux through the coil P is obtained to be equal to \[6.67\times {{10}^{-4}}\,Wb\].

So, the correct answer is “Option B”.

Note:
The units of the parameters should be taken care of. As, the given value of the magnetic flux is in weber, and in the options also, the unit of the magnetic flux is given to be weber. So, no, need for unit conversion from Tesla to weber or vice – versa.