CBSE Class 10 Physics Magnetic Effects of Electric Current MCQs Set 33

Answer: (c) current and magnetic field respectively

Explanation:
1. Fleming's left-hand rule: forefinger gives the field, middle finger gives the current and thumb gives the force.
2. So the middle finger and forefinger indicate current and magnetic field respectively.

Teacher's Note:
1. Remember: Fore = Field, Middle = Current, thumb = Thrust (force).
2. The three are mutually perpendicular.

Answer: (d) is useful in deciding the direction of an induced emf

Explanation:
1. Faraday's law gives the magnitude of the induced emf.
2. Lenz's law gives its direction: it opposes the change causing it.

Teacher's Note:
1. Direction: Lenz. Magnitude: Faraday.

Answer: (b) Fleming's left-hand rule

Explanation:
1. Fleming's left-hand rule is used for the motor effect (force on a conductor).
2. The right-hand rule gives the direction of induced current.

Teacher's Note:
1. Left hand for motors, right hand for generators.
2. Lenz's rule gives the direction of induced emf.

Answer: (c) zero

Explanation:
1. Force on a charge is F = qvB sin θ.
2. For a stationary charge v = 0, so the force is zero.

Teacher's Note:
1. A magnetic field acts only on moving charges.
2. The force depends on speed.

Answer: (d) a unit north pole

Explanation:
1. Magnetic field intensity at a point is the force on a unit north pole placed there.

Teacher's Note:
1. Electric field uses a unit positive charge.
2. Magnetic field uses a unit north pole.

Answer: (d) the fields of the magnet and the earth are equal and opposite

Explanation:
1. At a neutral point the net field is zero.
2. So the field of the magnet and that of the earth are equal in magnitude and opposite in direction.

Teacher's Note:
1. Neutral point: resultant field is zero.
2. Neither field is zero by itself.

Answer: (b) 0.06 N

Explanation:
1. F = BIL sin 90° = 0.30 × 5.0 × 0.04 = 0.06 N.

Teacher's Note:
1. Force on a conductor: F = BIL sin θ.
2. For a perpendicular field, sin θ = 1.

Answer: (d) zero

Explanation:
1. F = qvB sin θ, and θ = 0° for motion parallel to the field.
2. So the force is zero.

Teacher's Note:
1. Motion along the field line feels no force.
2. Maximum force occurs at 90°.

Answer: (a) change the velocity of a charged particle

Explanation:
1. A magnetic force changes only the direction of motion, so the velocity changes.
2. It does no work, so speed and kinetic energy do not change, and it cannot stop the particle.

Teacher's Note:
1. Force is always perpendicular to velocity.
2. Direction changes, speed does not.

Answer: (d) go from the south pole to the north pole

Explanation:
1. Field lines are closed curves.
2. Outside the magnet they go from north to south, so inside they return from south to north.

Teacher's Note:
1. Field lines never end.
2. Inside a magnet they run from S to N.

Answer: (b) ampere

Explanation:
1. Current, whether AC or DC, is measured in amperes.

Teacher's Note:
1. Farad is the unit of capacitance.

Answer: (b) is constant and opposite to the direction of the inducing current

Explanation:
1. Induced emf = –L dI/dt is constant when dI/dt is constant.
2. By Lenz's law it opposes the increase, so it is opposite to the inducing current.

Teacher's Note:
1. Constant rate of change gives constant emf.
2. Lenz: opposes the cause.

Answer: (c) true

Explanation:
1. At the magnetic equator the field is horizontal and the angle of dip is 0°.
2. So the vertical component B sin δ = 0, and the statement is true.

Teacher's Note:
1. Dip is 0° at the equator and 90° at the poles.
2. Vertical component = B sin(dip).

Answer: (b) the north–south direction

Explanation:
1. The loop acts like a magnetic dipole.
2. Like a freely suspended magnet, it lines up with its axis along the earth's magnetic field, which is north–south.

Teacher's Note:
1. Loop behaves as a magnet.
2. A freely suspended magnet rests in the north–south direction.

Answer: (b) the centre of the coil

Explanation:
1. Every element of the loop contributes a field in the same direction at the centre, and the centre is closest to all of them.
2. So the field is strongest at the centre.

Teacher's Note:
1. Field lines through the loop are densest at its centre.
2. On the axis, the field decreases as we move away from the centre.

Answer: (c) Existence of null points in the magnetic field of a bar magnet

Explanation:
1. A bar magnet placed in the earth's field has neutral points where the two fields cancel.
2. This shows the earth has a magnetic field like a magnet.

Teacher's Note:
1. The compass needle itself shows the earth's field.
2. Neutral points are due to the earth's field.

Answer: (b) compress

Explanation:
1. Neighbouring turns carry current in the same direction, so they attract one another.
2. So the spring contracts (compresses).

Teacher's Note:
1. Parallel currents in the same direction attract.
2. This is the principle of the solenoid contracting.

Answer: (a) 1.76 × 10⁶ m/s

Explanation:
1. The magnetic force provides the centripetal force: evB = mv²/r, so v = eBr/m.
2. v = (1.6 × 10^–19 × 10^–3 × 0.01)/(9.1 × 10^–31) = 1.76 × 10⁶ m/s.

Teacher's Note:
1. Radius of path r = mv/(eB).
2. Use the charge-to-mass ratio of the electron, 1.76 × 10¹¹ C/kg.

Answer: (c) it is economical in transmission

Explanation:
1. AC can be stepped up to a high voltage by a transformer, which reduces current and so the I²R loss in transmission.
2. This makes AC economical for long distances.

Teacher's Note:
1. High voltage means low current and low loss.
2. Transformers work only on AC.

Answer: (a) π

Explanation:
1. F = qvB sin θ, which is zero for θ = 0 or θ = π.
2. Among the options only π gives sin θ = 0.

Teacher's Note:
1. Zero force along or against the field.
2. Maximum force at π/2.

Answer: (c) true

Explanation:
1. At the magnetic poles the field is vertical, with a dip of 90°.
2. Horizontal component B cos 90° = 0, so the statement is true.

Teacher's Note:
1. At poles the compass needle stands vertical.
2. Horizontal component = B cos(dip).

Answer: (b) the composition of the wire

Explanation:
1. Motional emf e = Blv sin θ.
2. It depends on length, speed and orientation, but not on the material of the wire.

Teacher's Note:
1. Material affects resistance, not emf.