Unit 10: Electromagnetism — Short Questions

Lenz's law states that the direction of the induced current is such that it opposes the change in the magnetic flux that induces it.
It does not apply directly to induced e.m.f., it simply implies to the direction of current flow that opposes the change in magnetic flux. (As e.m.f. is potential difference, it does not have direction like current does.)

No, an emf is not induced in the loop. Since the normal to the loop is parallel to the magnetic field, the magnetic flux through the loop is not changing.
The flux is given by
φ = BA cos(θ)
Where θ is the angle between the normal and the magnetic field. Since θ = 0,
φ = BA, which is constant. Therefore,
Δφ = 0
ε = Δφ/Δt = 0, and ε = 0.

No, the induced emf does not always act to decrease the magnetic flux. According to Lenz's law, the induced emf acts to oppose the change in the magnetic flux. If the flux is increasing, the induced emf will act to decrease it. If the flux is decreasing, the induced emf will act to increase it.

The current is produced due to the changing magnetic flux through the solenoid. As the magnet is pushed into the solenoid, the magnetic flux increases, inducing an emf and a current. The direction of the current is determined by Lenz's law. The magnetic pole produced at the left end of the solenoid will be the same as the pole of the magnet being pushed into it (i.e., north).

No, the magnet will not fall with an acceleration of a freely falling body. As the magnet falls through the ring, it induces an emf and a current in the ring. The induced current produces a magnetic field that opposes the motion of the magnet, slowing it down. It is possible only when the force on ring towards magnet acts as south pole. As the gravitational force and magnetic force are opposite so net force acting on magnet is less than gravitational force which causes the acceleration less than gravitational acceleration.

The fast-moving charged particle will be deflected less than the slow-moving one. The force on a charged particle in a magnetic field is given by F = qvB sinθ, where v is the velocity of the particle. The radius of the circular path is given by r = mv/(qB).
For a given magnetic field and charge, the radius is directly proportional to the velocity. i.e r ∝ v
Therefore, the slow-moving particle will have a smaller radius and be deflected more.

To determine which particle will suffer greater deflection, we need to consider the force exerted on each particle by the magnetic field. The force on a charged particle in a magnetic field is given by the Lorentz force equation:
F = qvB sin(θ)
Where F is the force, q is the charge, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since both particles are projected at right angles to the magnetic field, θ = 90°, and sin (90°) = 1
Therefore, the force on each particle is:
F = qvB the direction of the force is perpendicular to both the velocity and the magnetic field. The radius of the circular path followed by each particle is given by:
r = mv / (qB)
Where m is the mass of the particle. For a given velocity and magnetic field, the radius is inversely proportional to the charge-to-mass ratio (q/m). The electron has a much larger q/m ratio than the proton (approximately 1836 times larger). Therefore, the electron will have a smaller radius and be deflected more than the proton.

Yes, a single moving proton can produce a magnetic field. A moving charge produces a magnetic field, and a proton is a charged particle. The magnetic force produced by a moving charge is given by
F = q v × B
Where B is the magnetic force, q is the charge, v is the velocity.

A magnetic field is necessary for a magnetic flux to pass through a coil, but it's not sufficient on its own. The magnetic flux through a coil is given by:
φ = B · ΔA
where φ is the magnetic flux, B is the magnetic field, and ΔA is the area element of the coil. For a magnetic flux to pass through a coil, the magnetic field lines must pass through the coil at some angle to the surface of coil not parallel to surface of coil.

According to right hand rule, the magnetic field lines are anti clock wise around the current carrying wire.

We know that the moving electric charge is similar to that of electric current. So it produces a magnetic field represented by the magnetic lines of force.

It is the arrangement consisting of electric and magnetic fields acting perpendicular to each other. The magnitudes of electric and magnetic fields and their directions are so adjusted that electric and magnetic force on the charged particle passing through it become equal. The particle passing through it suffers no deviation.
Mathematically we can write
Bqv = qE

No it is not possible to make a magnet having north or south pole. The existence of the pole is due to the currents, within an atom due to its spin and orbital motion. This is just like a current in a loop of wire, making its one side north and other side south pole.

The average e.m.f. induced in a coil of N turns is equal to the negative of the time rate of change of magnetic flux through the coil.
Mathematically: It is written as:
ε = -N Δφ/Δt
The negative sign gives the direction of e.m.f. which is such that it opposes the change of flux.

a) The emf induced across the ends of wire when it is moving in a magnetic field is called motional emf.
b) Its magnitude is written as
ε = Bv l Sinθ
Where B = magnetic induction
v = velocity of wire
l = length of wire
θ = The angle between velocity and magnetic induction.

No, a magnetic force is not exerted on a stationary charge. Magnetic force is exerted on moving charges. The expression for magnetic force on a charged particles is F = q (V × B) since charge is stationary
So, v = 0 hence
F = q (0 × B )
F = 0
The magnetic force on the charge is zero.

When the switch is closed, a current flows through the coil, generating a magnetic field. The changing magnetic field induces an electromotive force (emf) in the metal ring, causing it to jump upward due to the repulsive force between the coil and the ring. If the battery polarity is reversed, the direction of the current in the coil reverses, but the ring will still jump upward because the induced emf and the resulting force direction depend on the change in the magnetic field, not its direction.

To generate an emf, the coil must be rotated such that the magnetic flux through it changes. Rotating the coil around the x-axis will change the magnetic flux, but rotating around the y-axis or z-axis will not change because the angle between the coil's area and the magnetic field remains the same. Thus, rotation around the x-axis will generate an emf.

Yes, it is possible to change both area of loop and magnetic field without producing of emf. As we know
ε = -N Δφ/Δt
φ = (B.A)
If magnetic field is increased and area is decreased, such that product φ = B.A remains same, then there is no change of flux and therefore no emf is induced. Mathematically Δφ = 0, ε = 0

As magnetic force is perpendicular to v, so angle between F and v is also 90°
Hence, W = Fd cos90° as cos90° = 0
W = 0
Since no work is done. Thus, the kinetic energy remains unchanged.

The proton being positively charged will experience a force in the direction of electric field (opposite to its velocity) This will cause the proton to decelerate. The magnetic force of the proton will be perpendicular to both the magnetic field and velocity of proton. Since the velocity of proton is opposite to the magnetic field.
The proton will experience a deceleration due to electric field, but the magnetic field will not affect its motion then magnetic force will be zero (F = qvB sin(θ)) as θ = 180°.
F = qvBsin(180°)
F = qvB(0) = 0
So, F = 0

Yes, when a conductor is moved across a magnetic field, an electromotive force (emf) is induced in the conductor, which drives an electric current in the conductor

The electrical energy consumed by the bulb comes from the work done in moving the rod through the magnetic field.

See Q.3 of theory.

See Q.2 of theory.

See Q.8, 9 and 11 of theory.

See Q.5 of theory.

See Q.6 of theory.

See Q.12 of theory.