In a coil, the current changes form –2 A to +2A in 0.2 s and induces an emf of 0.1 V. The self-inductance of the coil is :
- 5 mH
- 1 mH
- 2.5 mH
- 4 mH
The Correct Option is A
Approach Solution - 1
To determine the self-inductance of the coil, we can use the following formula that relates the induced electromotive force (emf) in a coil to its self-inductance and the rate of change of current through it:
\(E = -L \frac{\Delta I}{\Delta t}\)
Where:
- \(E\) is the induced emf (0.1 V in this case).
- \(L\) is the self-inductance of the coil.
- \(\Delta I\) is the change in current (-2 A to +2 A, so \(\Delta I = 2 - (-2) = 4 \, \text{A}\)).
- \(\Delta t\) is the time interval over which the current changes (0.2 s).
Rearranging the formula to solve for \(L\), we have:
\(L = -\frac{E \cdot \Delta t}{\Delta I}\)
Substitute the given values into this formula:
\(L = -\frac{0.1 \, \text{V} \cdot 0.2 \, \text{s}}{4 \, \text{A}}\)
\(L = -\frac{0.02}{4}\)
\(L = -0.005 \, \text{H}\)
Since inductance is conventionally given as a positive value, we take the absolute value:
\(L = 0.005 \, \text{H} = 5 \, \text{mH}\)
Thus, the self-inductance of the coil is 5 mH.
Let's analyze the options:
- 5 mH: Correct, matches calculated value.
- 1 mH: Incorrect.
- 2.5 mH: Incorrect.
- 4 mH: Incorrect.
Therefore, the correct answer is 5 mH.
Approach Solution -2
The induced emf in a coil is given by:\[(\text{Emf})_{\text{induced}} = -L \frac{di}{dt}\]
In terms of magnitude:
\[|\text{Emf}_{\text{induced}}| = \left| L \frac{di}{dt} \right|\]
Given:
\[|\text{Emf}_{\text{induced}}| = 0.1 \, \text{V}\]
\[\frac{di}{dt} = \frac{2 - (-2)}{0.2} = \frac{4}{0.2} = 20 \, \text{A/s}\]
Now, solve for \( L \):
\[0.1 = L \times 20\]
\[L = \frac{0.1}{20} = 0.005 \, \text{H} = 5 \, \text{mH}\]
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Faradays laws of induction Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.