The dimension of mutual inductance is :
The dimension of mutual inductance is :
- [ML2T–2A–1]
- [ML2T–3A–1]
- [ML2T–2A–2]
- [ML2T–3A–2]
The Correct Option is C
Solution and Explanation
The correct answer is (C) : [ML2T–2A–2]
∵ U = \(\frac{1}{2}\) Mi²
⇒ [M] = \(\frac{[U]}{[i^2]}\) = \(\frac{ML^2T^{-2}}{A^2}\)
= [ML2T–2A–2]
Top Questions on Electromagnetic induction
A coil of area A and N turns is rotating with angular velocity \( \omega\) in a uniform magnetic field \(\vec{B}\) about an axis perpendicular to \( \vec{B}\) Magnetic flux \(\varphi \text{ and induced emf } \varepsilon \text{ across it, at an instant when } \vec{B} \text{ is parallel to the plane of the coil, are:}\)
- JEE Main - 2025
- Physics
- Electromagnetic induction
- A rectangular metallic loop is moving out of a uniform magnetic field region to a field-free region with a constant speed. When the loop is partially inside the magnetic field, the plot of the magnitude of the induced emf \( (\varepsilon) \) with time \( (t) \) is given by: \includegraphics[width=1\linewidth]{3.png}
- JEE Main - 2025
- Physics
- Electromagnetic induction
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of $\frac{1}{\sqrt{2}}$ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of $10 \mathrm{~cm} / \mathrm{s}$, induced emf between points A and E is _______ mV.}
- JEE Main - 2025
- Physics
- Electromagnetic induction
- If \( B \) is magnetic field and \( \mu_0 \) is permeability of free space, then the dimensions of \( \frac{B}{\mu_0} \) is:
- JEE Main - 2025
- Physics
- Electromagnetic induction
- Consider a long straight wire of a circular cross-section (radius \( a \)) carrying a steady current \( I \). The current is uniformly distributed across this cross-section. The distances from the center of the wire's cross-section at which the magnetic field (inside the wire, outside the wire) is half of the maximum possible magnetic field, anywhere due to the wire, will be:
- JEE Main - 2025
- Physics
- Electromagnetic induction
Questions Asked in JEE Main exam
- The number of points of discontinuity of the function $ f(x) = \left\lfloor \frac{x^2}{2} \right\rfloor - \left\lfloor \sqrt{x} \right\rfloor, \quad x \in [0, 4], $ where $ \left\lfloor \cdot \right\rfloor $ denotes the greatest integer function, is:
- JEE Main - 2025
- Functions
- A lens of focal length 20 cm in air is made of glass with a refractive index of 1.6. What is its focal length when it is immersed in a liquid of refractive index 1.8?
- JEE Main - 2025
- Wave optics
- Two charges \( 7 \, \mu C \) and \( -4 \, \mu C \) are placed at \( (-7 \, \text{cm}, 0, 0) \) and \( (7 \, \text{cm}, 0, 0) \) respectively. Given, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \), the electrostatic potential energy of the charge configuration is:
- JEE Main - 2025
- Electrostatics and Potential Energy
- The area of the region enclosed by the curves \( y = x^2 - 4x + 4 \) and \( y^2 = 16 - 8x \) is:
- JEE Main - 2025
- Area between Two Curves
- The ratio of intensities of two coherent sources is 1:9. The ratio of the maximum to the minimum intensities is:
- JEE Main - 2025
- Wave optics
Concepts Used:
Electromagnetic Induction
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
- When we place the conductor in a changing magnetic field.
- When the conductor constantly moves in a stationary field.
Formula:
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where
- e = induced voltage
- N = number of turns in the coil
- Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
- t = time
Applications of Electromagnetic Induction
- Electromagnetic induction in AC generator
- Electrical Transformers
- Magnetic Flow Meter