The induced emf in a coil rotating in a uniform magnetic field depend upon.
The correct answer is: All of the above
Lenz’s Law is due to conservation of :
The correct answer is: energy
A conducting rod is moved with a constant velocity v in a magnetic field. A potential difference appears across the two ends.
The correct answer is: None of these
A metallic rod completes its circuit as shown in the figure. The circuit is normal to a magnetic field of B = 0.15 tesla. If the resistances of the rod is 3Ω the force required to move the rod with a constant velocity of 2 m/sec is -
A coil of metal wire is kept stationary in a non-uniform magnetic field :
The correct answer is: neither emf nor current is induced
The magnetic flux φ (in weber) in a closed circuit of resistance 10Ω varies with time t (in second) according to equation φB = 6t2 - 5t + 1. The magnitude of induced current at t = 0.25sec is :
According to faraday’s Law.
If the flux of magnetic field across the coil varies with time an emf is induced in it.
The induced emf is equal to negative rate of change of magnetic flux across it.
Magnitude of induced emf = 2V, magnitude of induced current i = 0.2 A
The correct answer is: 0.2A
A coil has an area of 0.05 m2 and it has 800 turns. It is placed perpendicularly in a magnetic field of strength 4 × 10–5 wb/m2. It is rotated through 90° in 0.1s. The average emf induced in the coil is :
According to Faraday’s law
Where flux through the coil
θ is the angle between magnetic field and unit normal to the surface.
when coil is rotated from 0° to 90° then total change in magnetic flux is given by
= 0.016 V
The correct answer is: 0.016 V
A magnet is brought towards a coil (i) speedily and (ii) slowly then the induced emf/induced charge will be respectively.
According to Faraday’s law.
The magnitude of induced emf is directly proportional to the rate of change of magnetic flux. So induced emf will be more in first case.
We also know ε = iR
So, induced charge in a coil is independent of time taken to charge the flux.
The correct answer is: more in first case/equal in both case
The current carrying wire and rod AB are in the same plane. The rod moves parallel to the wire with a velocity v. Which one of the following statement is true about induced emf in the rod.
When the rod move in the magnetic field due to current carrying wire. The emf is induced in the rod.
Magnetic field due to current carrying wire is
So B is farther from A
So B is at lower potential.
The correct answer is: Potential at A will be higher than at B
The self inductance of solenoid of length L area of cross-section A and having N turns is
Let a solenoid cross-sectional area A and N turns. We know, in case of solenoid magnetic field along the axis of solenoid is given by
Where N is the number of turns and I is the current flowing in the solenoid.
Flux through coil of solenoid
Total flux = number of turns × flux through one coil.
Now we know
where L is self inductance