CBSE Class 10 Physics Magnetic Effects of Electric Current MCQs Set 30

Physics Objective Questions and Answers: Magnetic Effects of Electric Current

Access targeted multiple-choice questions for Magnetic Effects of Electric Current designed to align with the latest CBSE academic syllabus for Class 10 Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: A bar magnet is cut equally lengthwise and widthwise into four equal pieces. The pole strength of each piece compared with that of the original magnet is:
(a) same
(b) 1/8
(c) 1/4
(d) 1/2
Show Answer & Explanation

Answer: (d) 1/2

Explanation:
1. Cutting widthwise (across the length) leaves the cross-section, and so the pole strength, unchanged.
2. Cutting lengthwise halves the cross-section, so the pole strength halves.
3. Each piece therefore has half the original pole strength.

Teacher's Note:
1. Pole strength is proportional to the area of cross-section.
2. Poles can never be separated; each piece shows two new poles.

Question 2: Magnetic lines of force inside a bar magnet:
(a) run from the north pole to the south pole
(b) do not exist
(c) depend on the area of cross-section of the magnet
(d) run from the south pole to the north pole
Show Answer & Explanation

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

Explanation:
1. Outside the magnet lines run from N to S.
2. They are closed curves, so inside the magnet they run from S to N.

Teacher's Note:
1. Magnetic field lines form closed loops.

Question 3: The S.I. unit of magnetic field intensity (magnetic induction) is:
(a) weber
(b) tesla
(c) oersted
(d) gauss
Show Answer & Explanation

Answer: (b) tesla

Explanation:
1. The SI unit of B is the tesla (T = Wb/m²).
2. Gauss is the CGS unit.

Teacher's Note:
1. Weber is the unit of flux.

Question 4: The C.G.S. unit of magnetic field intensity is:
(a) tesla
(b) gauss
(c) weber
(d) none of these
Show Answer & Explanation

Answer: (b) gauss

Explanation:
1. The CGS unit of B is the gauss.
2. 1 T = 10⁴ gauss.

Teacher's Note:
1. Tesla is the SI unit.

Question 5: Two thin long parallel wires separated by a distance d carry a current i in the same direction. They will:
(a) attract each other
(b) repel each other
(c) depend on the material of the wires
(d) cannot say
Show Answer & Explanation

Answer: (a) attract each other

Explanation:
1. Each wire is in the field of the other.
2. By Fleming's left-hand rule, parallel currents attract.

Teacher's Note:
1. Parallel currents attract, antiparallel currents repel.

Question 6: Which rule gives the direction of the magnetic field due to a current-carrying conductor?
(a) Faraday's laws of electromagnetic induction
(b) Fleming's left-hand rule
(c) Lenz's law
(d) Maxwell's cork screw rule
Show Answer & Explanation

Answer: (d) Maxwell's cork screw rule

Explanation:
1. The cork-screw (right-hand thumb) rule gives the direction of the field of a current.
2. Fleming's left-hand rule gives the force.

Teacher's Note:
1. Distinguish: field direction by right-hand thumb rule, force by Fleming's left-hand rule.

Question 7: A transmission line carries a current of 100 A from east to west. The magnitude of the magnetic field due to this current at a distance 1 m below the wire is (μ₀ = 4π × 10⁻⁷ T m/A):
(a) 2 × 10⁻⁵ T
(b) 2 × 10⁵ T
(c) 10⁵ T
(d) 10⁻⁵ T
Show Answer & Explanation

Answer: (a) 2 × 10⁻⁵ T

Explanation:
1. \( B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{4\pi\times10^{-7}\times100}{2\pi\times1} \).
2. \( B = 2\times10^{-5} \) T.

Teacher's Note:
1. Field of a long straight wire is inversely proportional to distance.

Question 8: A proton is projected towards the east in a magnetic field directed towards the south. The proton will be deflected:
(a) towards north
(b) down
(c) up
(d) towards west
Show Answer & Explanation

Answer: (b) down

Explanation:
1. Take east = x, north = y, up = z. Then v = +x and B = −y.
2. \( \vec v\times\vec B = \hat x\times(-\hat y) = -\hat z \), i.e. downward.
3. For a positive charge the force is along \( \vec v\times\vec B \): downward.

Teacher's Note:
1. Fleming's left-hand rule: first finger field, middle finger current (motion of positive charge), thumb force.

Question 9: A circular conducting loop of radius r carries a constant current i and is placed in a uniform magnetic field B₀ perpendicular to the plane of the loop. The magnetic force on the loop is:
(a) \( i\pi r B_0 \)
(b) \( 2\pi r i B_0 \)
(c) zero
(d) \( i\pi r^2 B_0 \)
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. The forces on opposite elements of the loop are equal and opposite.
2. The net force on a closed current loop in a uniform field is zero (only a torque can exist).

Teacher's Note:
1. Closed loop, uniform field: net force zero.

Question 10: By inserting an iron core in a coil carrying current, the strength of its magnetic field will:
(a) increase
(b) decrease
(c) remain the same
(d) become zero
Show Answer & Explanation

Answer: (a) increase

Explanation:
1. Soft iron gets magnetised in the coil's field and adds its own field.
2. So the field becomes much stronger.

Teacher's Note:
1. This is how an electromagnet is made.

Question 11: Materials suitable for making electromagnets should have:
(a) high retentivity and high coercivity
(b) low retentivity and low coercivity
(c) high retentivity and low coercivity
(d) low retentivity and high coercivity
Show Answer & Explanation

Answer: (b) low retentivity and low coercivity

Explanation:
1. An electromagnet must magnetise strongly and lose magnetism when the current is switched off.
2. This needs low retentivity and low coercivity, as in soft iron.

Teacher's Note:
1. Permanent magnets need high retentivity and high coercivity.

Question 12: A proton moves with velocity 10⁴ m/s parallel to a magnetic field of 5 tesla. The force on the proton is:
(a) 8 × 10⁻¹⁵ N
(b) 10⁴ N
(c) 1.6 × 10⁻¹⁹ N
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. \( F = qvB\sin\theta \) with \( \theta = 0^\circ \).
2. So \( F = 0 \).

Teacher's Note:
1. No force when the motion is along the field.

Question 13: The angle of dip at the equator is:
(a) 0°
(b) 45°
(c) 60°
(d) 90°
Show Answer & Explanation

Answer: (a) 0°

Explanation:
1. At the magnetic equator the field is horizontal.
2. So the dip angle is 0°.

Teacher's Note:
1. At the poles the dip is 90°.

Question 14: The phenomenon of electromagnetic induction is:
(a) the process of charging a sphere
(b) the process of producing a magnetic field in a coil
(c) the process of producing an induced current in a coil whenever there is relative motion between the coil and a magnet
(d) the process of producing a cooling effect
Show Answer & Explanation

Answer: (c) the process of producing an induced current in a coil whenever there is relative motion between the coil and a magnet

Explanation:
1. Faraday showed that a changing magnetic flux induces an emf.
2. Relative motion of coil and magnet changes the flux.

Teacher's Note:
1. Induced emf needs a change of flux.

Question 15: The induced emf in a coil does NOT depend on:
(a) the number of turns in the coil
(b) the intensity of the magnetic field
(c) the relative speed between the coil and the magnet
(d) the resistance of the coil
Show Answer & Explanation

Answer: (d) the resistance of the coil

Explanation:
1. \( \varepsilon = -N\,\dfrac{\Delta\Phi}{\Delta t} \) depends on N, B and the rate of change.
2. Resistance affects only the current, not the emf.

Teacher's Note:
1. Current = emf ÷ resistance.

Question 16: A magnetic field of flux density 1.0 Wb m⁻² acts normal to a coil of 80 turns and area 0.01 m². The emf induced if the coil is removed from the field in 0.1 s is:
(a) 8 V
(b) 4 V
(c) 10 V
(d) 6 V
Show Answer & Explanation

Answer: (a) 8 V

Explanation:
1. Initial flux \( \Phi = BA = 1.0\times0.01 = 0.01 \) Wb; final flux 0.
2. \( \varepsilon = N\Delta\Phi/\Delta t = 80\times0.01/0.1 = 8 \) V.

Teacher's Note:
1. Multiply the flux change by the number of turns.

Question 17: A wire 88 cm long is bent into a circular loop and placed perpendicular to a magnetic field of 2.5 Wb m⁻². Within 0.5 s the loop is reshaped into a square and the flux density is increased to 3.0 Wb m⁻². The emf induced is approximately (take π = 22/7):
(a) 0.018 V
(b) 0.016 V
(c) 0.020 V
(d) 0.012 V
Show Answer & Explanation

Answer: (a) 0.018 V

Explanation:
1. Circle: \( 2\pi r = 88 \) cm gives \( r = 14 \) cm, area = 616 cm² = 0.0616 m².
2. Initial flux = 2.5 × 0.0616 = 0.154 Wb.
3. Square side = 22 cm, area = 484 cm² = 0.0484 m²; final flux = 3.0 × 0.0484 = 0.1452 Wb.
4. \( \varepsilon = (0.154-0.1452)/0.5 = 0.0176 \approx 0.018 \) V.

Teacher's Note:
1. A fixed perimeter encloses a larger area as a circle than as a square.

Question 18: A magnetic field of flux density 10 T acts normal to a coil of 50 turns and area 100 cm². The emf induced if the coil is removed from the field in 0.1 s is:
(a) 50 V
(b) 60 V
(c) 80 V
(d) 40 V
Show Answer & Explanation

Answer: (a) 50 V

Explanation:
1. Flux = 10 × 0.01 = 0.1 Wb.
2. \( \varepsilon = 50\times0.1/0.1 = 50 \) V.

Teacher's Note:
1. 100 cm² = 0.01 m².

Question 19: A transformer is a device which gives:
(a) DC voltage
(b) AC voltage
(c) both AC and DC voltage
(d) none of these
Show Answer & Explanation

Answer: (b) AC voltage

Explanation:
1. A transformer works on changing flux, so it needs AC input and gives AC output.

Teacher's Note:
1. It does not work with steady DC.

Question 20: A transformer is used:
(a) to transform electrical energy into mechanical energy
(b) to obtain suitable DC voltage
(c) to transform AC into DC
(d) to obtain a suitable AC voltage
Show Answer & Explanation

Answer: (d) to obtain a suitable AC voltage

Explanation:
1. Transformers step AC voltages up or down.

Teacher's Note:
1. Rectifiers change AC to DC.

Question 21: In a step-down transformer, which quantity is increased in the secondary?
(a) voltage
(b) current
(c) wattage
(d) none of these
Show Answer & Explanation

Answer: (b) current

Explanation:
1. Step down means lower voltage in the secondary.
2. Power is nearly the same, so current in the secondary is larger.

Teacher's Note:
1. V × I stays nearly constant.

Question 22: The principle of a transformer is:
(a) mutual induction
(b) self-induction
(c) electromagnetic induction alone
(d) none of these
Show Answer & Explanation

Answer: (a) mutual induction

Explanation:
1. The changing flux produced by the primary induces an emf in the neighbouring secondary coil.
2. This is mutual induction.

Teacher's Note:
1. Self-induction involves a single coil.

Question 23: The voltage induced in the secondary coil of a transformer is mainly due to:
(a) the iron core of the transformer
(b) the vibrations of the primary coil
(c) a varying induced electric field
(d) a varying magnetic flux (field) linked with it
Show Answer & Explanation

Answer: (d) a varying magnetic flux (field) linked with it

Explanation:
1. The alternating current in the primary produces an alternating magnetic flux in the core.
2. This changing flux through the secondary induces the emf.

Teacher's Note:
1. Without changing flux there is no induced emf.

Question 24: A transformer steps up an AC supply from 220 V to 2200 V. If the secondary coil has 2000 turns, the number of turns in the primary is:
(a) 200
(b) 100
(c) 50
(d) 20
Show Answer & Explanation

Answer: (a) 200

Explanation:
1. \( N_p/N_s = V_p/V_s = 220/2200 = 1/10 \).
2. \( N_p = 2000/10 = 200 \).

Teacher's Note:
1. Voltage ratio equals turns ratio.

Question 25: A step-down transformer reduces 220 V to 11 V. The primary draws 5 A and the secondary supplies 90 A. The efficiency of the transformer is:
(a) 4.4%
(b) 20%
(c) 33%
(d) 90%
Show Answer & Explanation

Answer: (d) 90%

Explanation:
1. Input power = 220 × 5 = 1100 W.
2. Output power = 11 × 90 = 990 W.
3. Efficiency = 990/1100 × 100 = 90%.

Teacher's Note:
1. Efficiency is output power divided by input power.

Question 26: In a transformer the primary has 140 turns and the secondary 280 turns. If the current in the primary is 4 A, the current in the secondary (ideal transformer) is:
(a) 4 A
(b) 2 A
(c) 6 A
(d) 10 A
Show Answer & Explanation

Answer: (b) 2 A

Explanation:
1. \( I_s/I_p = N_p/N_s = 140/280 \).
2. \( I_s = 4\times\tfrac12 = 2 \) A.

Teacher's Note:
1. Current ratio is the inverse of the turns ratio.

Question 27: According to Faraday's law, the total charge induced in a conductor moved in a magnetic field depends upon:
(a) the initial magnetic flux
(b) the final magnetic flux
(c) the rate of change of magnetic flux
(d) the change in magnetic flux
Show Answer & Explanation

Answer: (d) the change in magnetic flux

Explanation:
1. \( q = I\Delta t = \dfrac{\varepsilon}{R}\Delta t = \dfrac{\Delta\Phi}{R} \).
2. The charge depends on the total change of flux, not on how fast it changes.

Teacher's Note:
1. The rate of change affects the current, not the total charge.

Question 28: In an electric generator, the split rings (commutator) are used for:
(a) converting DC to AC
(b) converting AC to DC
(c) reducing power consumption
(d) splitting the current
Show Answer & Explanation

Answer: (b) converting AC to DC

Explanation:
1. The coil generates an alternating emf.
2. The split ring reverses the connection every half turn, so the external current is in one direction (DC).

Teacher's Note:
1. Slip rings are used in AC generators.

Question 29: In a simple DC motor the effect of using split rings is that:
(a) the direction of rotation of the coil is reversed
(b) the current in the coil always flows in the same direction
(c) the direction of the current in the coil is reversed every half turn
(d) none of these
Show Answer & Explanation

Answer: (c) the direction of the current in the coil is reversed every half turn

Explanation:
1. Without reversal the torque on the coil would reverse after half a turn.
2. The split ring reverses the current every half rotation, so the torque keeps the same sense.

Teacher's Note:
1. Commutator: reverses the current in the coil.

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