CBSE Class 10 Physics Magnetic Effects of Electric Current MCQs Set 35

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Chapter-wise Objective Questions: Magnetic Effects of Electric Current

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Multiple Choice Questions

Question 1: In Fleming's left-hand rule,
(a) the middle finger shows the direction of electric current.
(b) the fore finger shows the direction of motion of the conductor.
(c) the thumb shows the magnetic field.
(d) the middle finger and thumb are not mutually perpendicular to each other.
Show Answer & Explanation

Answer: (a) the middle finger shows the direction of electric current.

Explanation:
1. In Fleming's left-hand rule the fore finger, middle finger and thumb are mutually perpendicular.
2. The fore finger gives the magnetic field, the middle finger the current and the thumb the force (motion) on the conductor.

Teacher's Note:
1. Left hand rule is for motors (force on a current-carrying conductor).
2. Right-hand rule is for generators.

Question 2: N1 and N2 are the numbers of turns in the primary and secondary coils of a transformer. For a step-up transformer,
(a) N1 < N2
(b) N1 = N2
(c) N1 > N2
(d) insufficient information
Show Answer & Explanation

Answer: (a) N1 < N2

Explanation:
1. V2/V1 = N2/N1.
2. A step-up transformer raises the voltage, so N2 > N1.

Teacher's Note:
1. Step-up: more turns in the secondary.
2. Step-down: more turns in the primary.

Question 3: Which of the following is an incorrect statement?
(a) The angle of declination changes from place to place.
(b) The angle of inclination (dip) changes from place to place.
(c) The angle of dip is 90° at the magnetic poles.
(d) The angle of dip is minimum at the magnetic poles.
Show Answer & Explanation

Answer: (d) The angle of dip is minimum at the magnetic poles.

Explanation:
1. Dip is zero at the magnetic equator and 90° (maximum) at the magnetic poles.
2. So the statement that dip is minimum at the poles is incorrect.

Teacher's Note:
1. Dip: 0° at the equator, 90° at the poles.
2. Declination and dip both vary with location.

Question 4: The galvanometer constant is
(a) inversely proportional to the number of turns in the coil.
(b) directly proportional to the area of cross-section of the coil.
(c) directly proportional to the magnetic field.
(d) inversely proportional to the couple per unit twist.
Show Answer & Explanation

Answer: (a) inversely proportional to the number of turns in the coil.

Explanation:
1. Galvanometer constant K = C/(nBA), where C is the torsional couple per unit twist.
2. K is inversely proportional to n (and to B and A) and directly proportional to C.

Teacher's Note:
1. A sensitive galvanometer has a small K.
2. More turns give a larger deflection per unit current.

Question 5: A current of 5 A flows through a coil of 500 turns, each of radius π cm. By how much does the magnetic induction B at the centre change if the same wire is made into a coil of 250 turns?
(a) 3.75 × 10-2 T
(b) 3.75 × 10-2 μT
(c) 7.75 × 10-4 μT
(d) 3.75 T
Show Answer & Explanation

Answer: (a) 3.75 × 10-2 T

Explanation:
1. Case I: B = nμ0i/(2r) = 500 × 4π × 10-7 × 5/(2π × 10-2) = 5 × 10-2 T.
2. The wire length is fixed (ℓ = n × 2πr), so halving n doubles r to 2π cm.
3. Case II: B = 250 × 4π × 10-7 × 5/(2 × 2π × 10-2) = 1.25 × 10-2 T.
4. Change = 5 × 10-2 - 1.25 × 10-2 = 3.75 × 10-2 T.

Teacher's Note:
1. Keep the total wire length constant when the number of turns changes.
2. B at the centre = nμ0i/2r.

Question 6: 120 V AC is connected to a coil of self-inductance 20 H. What is the rate of increase of current?
(a) 0.166 A s-1
(b) 3 A s-1
(c) 6 A s-1
(d) 8 A s-1
Show Answer & Explanation

Answer: (c) 6 A s-1

Explanation:
1. Induced emf E = L(di/dt).
2. di/dt = E/L = 120/20 = 6 A s-1.

Teacher's Note:
1. Self-inductance opposes change in current.
2. Unit of L is the henry.

Question 7: The relative permeability is the least in ______ magnetic substances.
(a) dia
(b) ferro
(c) ferri
(d) para
Show Answer & Explanation

Answer: (a) dia

Explanation:
1. Diamagnetic substances have μr slightly less than 1.
2. Para and ferro have μr greater than 1, so dia is the least.

Teacher's Note:
1. Diamagnetic substances are weakly repelled by magnets.
2. Ferromagnetic substances have very large μr.

Question 8: A coil of area 20 cm2 is placed perpendicular to a magnetic field whose strength changes from 200 T to 400 T in 0.1 s. What emf (in V) is induced?
(a) 8
(b) 4
(c) 2
(d) 1
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Induced emf = A(ΔB)/Δt = 20 × 10-4 × 200/0.1 = 4 V.

Teacher's Note:
1. Convert cm2 to m2 (× 10-4).
2. Faraday's law: emf = rate of change of flux.

Question 9: The primary and secondary coils of a transformer have 200 and 300 turns respectively. What is the output voltage for an input of 200 V?
(a) 1200 V
(b) 900 V
(c) 300 V
(d) 600 V
Show Answer & Explanation

Answer: (c) 300 V

Explanation:
1. Vs/Vp = Ns/Np.
2. Vs = 200 × 300/200 = 300 V.

Teacher's Note:
1. It is a step-up transformer.
2. Output current is smaller than input current.

Question 10: Two infinitely long straight wires carry currents of 20 A (upwards) and 30 A (downwards), and the point P lies in the plane of the wires midway between them, 5 cm from each. What is the resultant magnetic field at P?
(a) 5 × 10-4 T
(b) 8 × 10-2 T
(c) 2 × 10-4 T
(d) 10-3 T
Show Answer & Explanation

Answer: (c) 2 × 10-4 T

Explanation:
1. B1 = (μ0/2π)(i/r) = 2 × 10-7 × 20/0.05 = 8 × 10-5 T.
2. B2 = 2 × 10-7 × 30/0.05 = 1.2 × 10-4 T.
3. For opposite currents the fields between the wires point in the same direction, so B = 8 × 10-5 + 1.2 × 10-4 = 2 × 10-4 T.

Teacher's Note:
1. Between antiparallel currents the fields add.
2. Use the right-hand grip rule for direction.

Question 11: What is the magnetic field at the centre of a circular coil of radius 5 cm carrying a current of 50 A (one turn)?
(a) 2π × 10-4 T
(b) 3π × 10-6 T
(c) π × 10-6 T
(d) 4π × 10-17 T
Show Answer & Explanation

Answer: (a) 2π × 10-4 T

Explanation:
1. B = μ0i/(2r) = 4π × 10-7 × 50/(2 × 0.05).
2. B = 2π × 10-4 T.

Teacher's Note:
1. At the centre of a loop B = μ0i/2r.
2. The field is along the axis of the loop.

Question 12: A current-carrying conductor of mass 20 g and length 50 cm lies perpendicular to a magnetic field of 20 T on a frictionless horizontal surface. Arrange the steps to find the acceleration when 2 A flows.
(A) Note the given data.
(B) Use F = Bil.
(C) Use F = ma.
(D) Equate the two expressions.
(E) Determine the acceleration.
(a) A B C D E
(b) B C D A E
(c) A C D B E
(d) B C A D E
Show Answer & Explanation

Answer: (a) A B C D E

Explanation:
1. Note the data (A), write the magnetic force F = Bil (B) and Newton's law F = ma (C).
2. Equate them (D) to get a = Bil/m = 20 × 2 × 0.5/0.02 = 1000 m s-2 (E).

Teacher's Note:
1. Magnetic force on a conductor is Bil sinθ.
2. Convert mass to kg before substituting.

Question 13: Assertion (A): An iron rod is used to make a temporary magnet.
Reason (R): The susceptibility of iron is high and positive.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) Both A and R are false.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Iron has low retentivity, so it loses its magnetism easily and suits a temporary magnet, so A is true.
2. Iron's susceptibility is indeed high and positive, so R is true, but the reason for A is the low retentivity, so R is not the explanation.

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

Question 14: At what angle between the conductor and the magnetic field is the magnetic force on a current-carrying conductor maximum?
(a) 0°
(b) 30°
(c) 90°
(d) 45°
Show Answer & Explanation

Answer: (c) 90°

Explanation:
1. F = Bil sinθ is maximum when sinθ = 1, i.e. θ = 90°.

Teacher's Note:
1. The force is zero when the conductor is parallel to the field.

Question 15: For a step-down transformer,
(a) the input voltage is greater than the output voltage.
(b) the input current is greater than the output current.
(c) the primary voltage is stepped up to a higher value.
(d) the number of turns of the primary coil is less than that of the secondary coil.
Show Answer & Explanation

Answer: (a) the input voltage is greater than the output voltage.

Explanation:
1. A step-down transformer lowers the voltage, so the input voltage exceeds the output voltage.
2. Its primary has more turns than its secondary, and the output current is greater than the input current.

Teacher's Note:
1. Power is nearly the same on both sides.
2. Lower voltage means higher current.

Question 16: Which of the following is an incorrect statement?
(a) The angle of dip is maximum at the poles.
(b) The angle of dip is minimum at the equator.
(c) The angle of dip is maximum at the equator.
(d) At the magnetic equator of the Earth the angle of dip is zero.
Show Answer & Explanation

Answer: (c) The angle of dip is maximum at the equator.

Explanation:
1. The dip is 90° (maximum) at the magnetic poles and 0° (minimum) at the magnetic equator.
2. So the statement that the dip is maximum at the equator is incorrect.

Teacher's Note:
1. Dip needle is horizontal at the equator.

Question 17: A galvanometer becomes
(a) a voltmeter if a high resistance is connected in series with it.
(b) an ammeter if a low resistance is connected in parallel with it.
(c) an ammeter if a high resistance is connected in series with it.
(d) both (a) and (b).
Show Answer & Explanation

Answer: (d) both (a) and (b).

Explanation:
1. A large series resistance converts a galvanometer into a voltmeter.
2. A small shunt resistance in parallel converts it into an ammeter.
3. So (a) and (b) are both correct, and (c) is wrong.

Teacher's Note:
1. Ammeter: low resistance, in series with the circuit.
2. Voltmeter: high resistance, in parallel across the element.

Question 18: Currents of 2 A and 3 A flow through two coils, and the magnetic fields at their centres are 10 T and 15 T respectively. If the ratio of the numbers of turns is 1 : 2, what is the ratio of the radii of the coils?
(a) 1 : 3
(b) 1 : 2
(c) 1 : 4
(d) 1 : 5
Show Answer & Explanation

Answer: (b) 1 : 2

Explanation:
1. B = nμ0i/(2r), so r ∝ ni/B.
2. r1/r2 = (n1i1/B1)/(n2i2/B2) = (1 × 2/10)/(2 × 3/15) = 0.2/0.4 = 1/2.

Teacher's Note:
1. Write r in terms of the other quantities and take a ratio.

Question 19: A 200 V AC supply is connected to a coil. If the current increases at the rate of 4 A s-1, what is the inductance of the coil?
(a) 50 H
(b) 60 H
(c) 0.02 H
(d) 55 H
Show Answer & Explanation

Answer: (a) 50 H

Explanation:
1. E = L(di/dt), so L = 200/4 = 50 H.

Teacher's Note:
1. The applied emf balances the back emf L di/dt.

Question 20: The magnetic susceptibility is the highest in ______ magnetic substances.
(a) dia
(b) ferro
(c) ferri
(d) para
Show Answer & Explanation

Answer: (b) ferro

Explanation:
1. Ferromagnetic materials such as iron have very large positive susceptibility.
2. Diamagnetic is small and negative; paramagnetic is small and positive.

Teacher's Note:
1. Susceptibility χ = I/H.

Question 21: A coil is placed perpendicular to a uniform magnetic field of 200 T. If the area of the coil changes from 20 cm2 to 25 cm2 in 0.1 s, what is the induced emf (in V)?
(a) 3
(b) 8
(c) 1
(d) 2
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. Induced emf = B(ΔA)/Δt = 200 × (5 × 10-4)/0.1 = 1 V.

Teacher's Note:
1. ΔA = 5 cm2 = 5 × 10-4 m2.

Question 22: A current of 20 A passes through the primary coil of a transformer. The ratio of turns in the primary to the secondary is 1 : 5. What is the output current?
(a) 10 A
(b) 4 A
(c) 8 A
(d) 16 A
Show Answer & Explanation

Answer: (b) 4 A

Explanation:
1. Np/Ns = Is/Ip, so Is = 20 × 1/5 = 4 A.

Teacher's Note:
1. Step-up transformers reduce the current.

Question 23: Two infinitely long parallel wires carry currents of 5 A and 20 A and a point P lies midway between them, 10 cm from each. What is the ratio of the magnetic field due to the 5 A wire to that due to the 20 A wire at P?
(a) 1/5
(b) 1/2
(c) 1/3
(d) 1/4
Show Answer & Explanation

Answer: (d) 1/4

Explanation:
1. B = (μ0/2π)(i/r), and r is the same for both wires.
2. BA/BB = 5/20 = 1/4.

Teacher's Note:
1. At equal distances B is proportional to the current.

Question 24: What current flows through a circular coil (one turn) if the magnetic field at its centre is 4 × 10-5 T and its radius is 4 cm?
(a) \( \frac{1}{\pi} \) A
(b) \( \frac{2}{\pi} \) A
(c) \( \frac{8}{\pi} \) A
(d) \( \frac{4}{\pi} \) A
Show Answer & Explanation

Answer: (c) \( \frac{8}{\pi} \) A

Explanation:
1. B = μ0i/(2r) gives i = 2rB/μ0.
2. i = 2 × 0.04 × 4 × 10-5/(4π × 10-7) = 8/π A.

Teacher's Note:
1. Keep all quantities in SI units.

Question 25: A coil of mass 30 g lies on a frictionless horizontal surface and a magnetic field is applied perpendicular to its wire, of length 1 m, carrying 5 A. The acceleration produced is 1 m s-2. Arrange the steps to find B.
(A) Use F = Bil.
(B) Note the given values.
(C) Use F = ma.
(D) Equate the two and find B.
(a) B D A C
(b) B A C D
(c) D C A B
(d) D A C B
Show Answer & Explanation

Answer: (b) B A C D

Explanation:
1. Note the data (B), write F = Bil (A) and F = ma (C), then equate them (D).
2. B = ma/(iℓ) = 0.03 × 1/(5 × 1) = 6 × 10-3 T.

Teacher's Note:
1. Mass must be in kg.

Question 26: Assertion (A): A body can possess a charge of 8 × 10-20 C.
Reason (R): The charge on a body is a sub-multiple of the charge of an electron.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) Both A and R are false.
Show Answer & Explanation

Answer: (d) Both A and R are false.

Explanation:
1. Charge is quantised: q = ne with e = 1.6 × 10-19 C.
2. 8 × 10-20 C = 0.5e is not an integer multiple, so A is false.
3. Charge is always an integral multiple of e, not a sub-multiple, so R is false.

Teacher's Note:
1. Charge is quantised.

Question 27: Match Column A with Column B.
(A) Magnetic field induction (B) 1 gauss (C) Intensity of magnetic field (D) Magnetic susceptibility
(P) A m-1 (Q) no unit (R) directly proportional to the current in the wire (S) 0.0001 SI units of magnetic induction
(a) A → Q; B → R; C → P; D → S
(b) A → S; B → R; C → Q; D → P
(c) A → P; B → Q; C → R; D → S
(d) A → R; B → S; C → P; D → Q
Show Answer & Explanation

Answer: (d) A → R; B → S; C → P; D → Q

Explanation:
1. B around a wire is directly proportional to the current (R).
2. 1 gauss = 10-4 tesla (S).
3. The SI unit of the intensity of magnetic field H is A m-1 (P).
4. Susceptibility is a ratio and has no unit (Q).

Teacher's Note:
1. 1 tesla = 104 gauss.
2. χ = I/H is dimensionless.

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