Practice free Electromagnetic Induction (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
The self inductance of a straight conductor is…
The self-inductance of a straight conductor is zero because self-inductance depends on the flux linkage with the coil formed by the conductor. A straight conductor does not form a coil, and hence, there is no flux linkage, resulting in zero self-inductance.
Two coils of self inductances 2 mH & 8 mH are placed so close togather that the effective flux in one coil is completely half with the other.The mutual inductance between these coils is.......
$ L = k \sqrt { L_0 L_2 } , k = 1/2 $
In circular coil. when no. of turns is doubled & resistance becomes half of the initial then inductance becomes
$ L \alpha N^2 $
The armature of dc motor has $ 20 \Omega $ resistance. It draws current of 1.5 A when run by 220 V dc suppy. The value of back induced in it will be
The back EMF (Electromotive Force) in a DC motor can be calculated using Ohm's Law and the formula for the back EMF. The voltage drop across the armature resistance is given by $V_R = I imes R$, where $I$ is the current and $R$ is the resistance. Given that the armature resistance is 20 Ω and the current is 1.5 A, the voltage drop is $V_R = 1.5 imes 20 = 30V$. The back EMF $E_b$ can be found by subtracting this voltage drop from the supply voltage: $E_b = V_{supply} - V_R = 220V - 30V = 190V$.
Two identical circular loops of metal wire are lying on a table near to each other without touching. Loop A carries a current which increasing with time. In response the loop B.......
When a current in loop A is increasing, it creates an increasing magnetic field. According to Lenz's Law, loop B will experience an induced current that opposes the change in magnetic flux. This induced current in loop B will create its own magnetic field, which will oppose the magnetic field of loop A. Therefore, loop B is repelled by loop A.
A wire of length 2m is moving at a speed $2ms^{-1}$ keep its length perpendicular to uniform magnetic field of 0.5 T. The resistance of circuit joined with this wire is $6 \Omega $ . The rate at which work is being done to keep the wire moving at constant speed is ..........
$ \rho = F.v. = {B^2 l^2 v^2 \over R } = {1 \over 6 } W $
An inductor-resistor-battery circuit is switched on at t = 0. If the emf of battry is $ \varepsilon $ find the charge passes through the battery in one time constant $ \tau $
$ I = I_0 (1 - e^{-t /T}) (I_0 =I_{max}) $ $ Q = \int _0 ^ T (I_0 -I_0e ^{-t /T })dt = I_0 T/e $
A metal rod of length 2m rotates vertically about one of its end with frequency 2 Hz. The horizontal component of earth's magnetic field is $3.14 \times 10^{-5} T$ then emf developed between two ends of road is.....
The self inductance of a coil is 5H, a current of 1A changes to 2Awithin 5 sec. through the coil. The value of induced emf will be.......
The induced electromotive force (emf) in a coil is given by Faraday's Law of Electromagnetic Induction, which states that:
\[ \text{emf} = -L \frac{dI}{dt} \]
Where:
Here, the current changes from 1A to 2A in 5 seconds. Therefore, \(\frac{dI}{dt} = \frac{2A - 1A}{5s} = \frac{1}{5} = 0.2 A/s\).
Substituting the values, we get:
\[ \text{emf} = -5H \times 0.2 A/s = -1 V \]
The negative sign indicates the direction of the induced emf. Thus, the magnitude of the induced emf is 1V.
A coil of inductance 300 mH and resistance $ 2 \Omega $ is connected to a source of voltage 2V. The current reaches half of its steady state value in.........
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every Electromagnetic Induction question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.