Question

A solenoid of 10 Henry inductance and 2-ohm resistance, is connected to a 10-volt battery. In how much time the magnetic energy will be reached to \[1/4\text{th}\]of the maximum value.

A.3.5 sec
B.2.5 sec
C.5.5 sec
D.7.5 sec

Answer 1

Hint: The net magnetic field is the sum from each individual loop and is maximum in the middle of the solenoid because that point minimizes the average distance to each loop


Complete step-by-step solution:
Given,
Inductance, \[L=10\text{ H}\]
Resistance, \[r=2\text{ ohm}\]
Volt, \[V=10\text{ V}\]

Since, \[V=iR\]; where i is current

Therefore, current, \[{{i}_{o}}=\dfrac{10}{2}=5\text{A}\]

Now Maximum energy is,
\[\begin{gathered}
  & {{E}_{0}}=\dfrac{1}{2}L\times {{i}_{o}} \\
 & \text{ }=\dfrac{1}{2}\left( 10 \right)\times {{\left( 5 \right)}^{2}} \\
 & \text{ }=125\text{ J} \\
\end{gathered}\]


Thus. \[1/4\text{th}\]of the total energy is,
\[\begin{gathered}
  & E=\dfrac{1}{4}{{E}_{0}} \\
 & \text{ }=\dfrac{{{E}_{0}}}{4} \\
 & \text{ }=\dfrac{125}{4}\text{ J} \\
\end{gathered}\]

When \[E=\dfrac{125}{4}\text{ J}\], assume the current to be \[i\]
Now,
Current \[i\] is,

\[\begin{gathered}
  & \text{ }E=\dfrac{1}{2}\times L\times {{i}^{2}} \\
 & \Rightarrow \dfrac{125}{4}=\dfrac{1}{2}\times \left( 10 \right)\times {{i}^{2}} \\
 & \Rightarrow \dfrac{125}{4}=5\times {{i}^{2}} \\
 & \Rightarrow i=\dfrac{5}{2} \\
 & \Rightarrow i=2.5\text{ A} \\
\end{gathered}\]

Now time taken to rise from 0 A to \[2.5\text{ A}\],

\[\begin{gathered}
  & \text{ }i={{i}_{o}}\left[ 1-{{e}^{-Rt/L}} \right] \\
 & \Rightarrow 2.5=5\left[ 1-{{e}^{-Rt/L}} \right] \\
 & \Rightarrow {{e}^{-Rt/L}}=0.5 \\
 & \Rightarrow \dfrac{-Rt}{L}={{\log }_{e}}\left( 0.5 \right) \\
 & \Rightarrow \dfrac{Rt}{L}=0.693 \\
 & \Rightarrow t=\dfrac{6.93}{2} \\
 & \Rightarrow t=3.46\,\approx 3.5\text{ sec} \\
\end{gathered}\]

Hence, the correct answer is option A.


Note: while solving the formula, \[i={{i}_{o}}\left[ 1-{{e}^{-Rt/L}} \right]\] make sure that the exponential function is converted to logarithm function to make the solution easier to solve.