Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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 increased to $\dfrac{1}{4}th$ of the maximum value.
A) 3.5 sec.
B) 2.5 sec.
C) 5.5 sec.
D) 7.5 sec.

seo-qna
Last updated date: 20th Jun 2024
Total views: 394.8k
Views today: 7.94k
Answer
VerifiedVerified
394.8k+ views
Hint:The inductance is defined as the tendency of the conductor to oppose the flow of current through it. Inductance is defined as the ratio of the induced voltage to change of current. The flow of current creates a magnetic flow around the inductor.

Formula used:The ohm's law is given by,
$ \Rightarrow V = I \times R$

Where potential difference is V, the current is I and the resistance is R.
The energy of inductance is given by,

$ \Rightarrow E = \dfrac{1}{2} \times L \times {i^2}$
Where inductance is L, the current is $i$.

Complete step by step solution:
It is given in the problem that a solenoid of 10 henry inductance and 2 ohm resistance is connected to a 10 volt battery and we need to find the time taken for the magnetic energy to increase to $\dfrac{1}{4}th$ of the maximum value.

First of all we need to find the maximum current in the circuit.

$ \Rightarrow {i_o} = \dfrac{V}{R}$
$ \Rightarrow {i_o} = \dfrac{{10}}{2}$
$ \Rightarrow {i_o} = 5A$.

The maximum magnetic energy can be calculated by using the formula. The energy of inductance is given by,

$ \Rightarrow {E_o} = \dfrac{1}{2} \times L \times i_{\max. }^2$

Where inductance is L, the maximum current is ${i_{\max. }}$.

$ \Rightarrow {E_o} = \dfrac{1}{2} \times L \times i_{\max. }^2$
$ \Rightarrow {E_o} = \dfrac{1}{2} \times 10 \times {\left( 5 \right)^2}$
$ \Rightarrow {E_o} = 5 \times 25$
$ \Rightarrow {E_o} = 125J$.

The magnetic energy will be increased to $\dfrac{1}{4}th$ of the maximum energy.

$ \Rightarrow E = \dfrac{{{E_o}}}{4}$
$ \Rightarrow E = \dfrac{{125}}{4}$
$ \Rightarrow E = \dfrac{{125}}{4}$

The energy of inductance is given by,

$ \Rightarrow E = \dfrac{1}{2} \times L \times {i^2}$

Where inductance is L, the current is $i$.

$ \Rightarrow E = \dfrac{1}{2} \times L \times {i^2}$
$ \Rightarrow \dfrac{{125}}{4} = \dfrac{1}{2} \times 10 \times {i^2}$
$ \Rightarrow \dfrac{{125}}{2} = 10 \times {i^2}$
$ \Rightarrow \dfrac{{125}}{{20}} = {i^2}$
$ \Rightarrow {i^2} = \dfrac{{125}}{{20}}$
$ \Rightarrow i = \sqrt {\dfrac{{125}}{{20}}} $
$ \Rightarrow i = 2 \cdot 5A$.

The current will become 2.5A.
The formula of the instantaneous current during its growth in L-R circuit is given by,

$ \Rightarrow i = {i_o}\left( {1 - {e^{ - \dfrac{{Rt}}{L}}}} \right)$

Where current is $i$, the maximum current is ${i_o}$, the resistance is R, the inductor is L and the time taken is t.

$ \Rightarrow i = {i_o}\left( {1 - {e^{ - \dfrac{{Rt}}{L}}}} \right)$
$ \Rightarrow 2 \cdot 5 = 5\left( {1 - {e^{ - \dfrac{{2 \times t}}{{10}}}}} \right)$
$ \Rightarrow 0 \cdot 5 = \left( {1 - {e^{ - \dfrac{t}{5}}}} \right)$
$ \Rightarrow {e^{ - \dfrac{t}{5}}} = 0 \cdot 5$

Taking logarithms on both sides.

$ \Rightarrow \ln \left( {{e^{\dfrac{{ - t}}{5}}}} \right) = \ln \left( {0 \cdot 5} \right)$
$ \Rightarrow \dfrac{{ - t}}{5}\ln \left( e \right) = \ln \left( {0 \cdot 5} \right)$
$ \Rightarrow \dfrac{t}{5} = - \ln \left( {0 \cdot 5} \right)$
$ \Rightarrow \dfrac{t}{5} = \ln \left( {0 \cdot {5^{ - 1}}} \right)$
$ \Rightarrow \dfrac{t}{5} = 0 \cdot 693$
$ \Rightarrow t = 5 \times 0 \cdot 693$
$ \Rightarrow t = 3 \cdot 465$
$ \Rightarrow t \approx 3 \cdot 5\sec $

The time taken by the magnetic energy to reach $\dfrac{1}{4}th$ of its value is $t = 3 \cdot 5\sec $.

The correct answer for this problem is option A.

Note:The students are advised to understand and remember the formula of the energy of inductance and also the formula of the ohm's law as these formulas are very useful in solving the problem like these. The inductor forms a magnetic field around the inductor which opposes the current which is flowing through the inductor.