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JEE Main Electricity & Magnetism: practice questions with solutions

13 questions on Alternating Current, Capacitors, Current Electricity, Electromagnetic Induction, Electrostatics, Magnetic Effects of Current, Moving Charges & Magnetism. Try each one first, then open the solution.

Tip: solve on paper before opening a solution. Once you open it, that question is counted as practised and won't appear in your HeyGyan tests.
ElectrostaticsHard

Q1. Two identical conducting spheres with charges +q and −3q attract each other with force F at separation r. They are touched together and then placed at separation r/2. What is the new force?

  1. 4F/3, repulsive
  2. 4F/3, attractive
  3. F/3, repulsive
  4. 3F/4, repulsive
Show answer and solution

Answer: (A) 4F/3, repulsive

SolutionInitially F = k(3q²)/r². After contact the total charge −2q splits equally: −q each, so they repel. New force = kq²/(r/2)² = 4kq²/r² = (4/3)F.
Common trapForgetting that the force turns repulsive once both spheres carry the same sign.
CapacitorsMedium

Q2. A parallel-plate capacitor is charged and then disconnected from the battery. A dielectric slab of K = 4 is then inserted to fill the gap completely. The stored energy becomes:

  1. 4 times
  2. 2 times
  3. one-half
  4. one-fourth
Show answer and solution

Answer: (D) one-fourth

SolutionOnce disconnected, charge Q stays constant. Capacitance becomes 4C. Energy U = Q²/2C, so it falls to one-fourth.
Common trapUsing U = ½CV² with constant V. That only applies while the battery stays connected.
Magnetic Effects of CurrentMedium

Q3. A circular loop carrying current I produces field B at its centre. The same wire is rewound into a coil of 2 turns carrying the same current. What is the field at the centre now?

  1. B/2
  2. 2B
  3. 4B
  4. 8B
Show answer and solution

Answer: (C) 4B

SolutionSame wire length, so 2 turns means radius R/2. B = μ₀nI/2r. With n doubled and r halved, B becomes 2 × 2 = 4 times.
Common trapCounting only the extra turn (2B) and missing that the radius also halves.
Electromagnetic InductionMediumNumerical

Q4. A metal rod 1 m long rotates at 20 rad/s about one end, in a plane perpendicular to a uniform magnetic field of 0.5 T. What emf (in volts) is induced between its ends?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 5

SolutionFor a rod rotating about one end, emf = ½BωL² = ½ × 0.5 × 20 × 1² = 5 V.
Common trapUsing BωL² without the ½. The speed varies along the rod, so its average is ωL/2.
Current ElectricityHard

Q5. Twelve equal resistors R form the edges of a cube. What is the equivalent resistance between two diagonally opposite corners of the cube?

  1. R/2
  2. 7R/12
  3. 3R/4
  4. 5R/6
Show answer and solution

Answer: (D) 5R/6

SolutionBy symmetry, current splits 3 ways, then 6 ways, then 3 ways. With current I: drop = (I/3)R + (I/6)R + (I/3)R = 5IR/6. So R_eq = 5R/6.
Common trap7R/12 is for adjacent corners and 3R/4 for a face diagonal.
Alternating CurrentEasyNumerical

Q6. A series LCR circuit has L = 0.1 H and C = 10 μF. What is its resonant angular frequency (in rad/s)?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 1000

Solutionω₀ = 1/√(LC) = 1/√(0.1 × 10 × 10⁻⁶) = 1/√(10⁻⁶) = 1000 rad/s.
Common trapDividing by 2π gives frequency in Hz (about 159 Hz), not angular frequency.
Moving Charges & MagnetismHard

Q7. A proton and an alpha particle with equal kinetic energies enter a uniform magnetic field perpendicular to it. What is the ratio of the radius of the proton's path to that of the alpha particle?

  1. 1 : 1
  2. 1 : 2
  3. 2 : 1
  4. 1 : √2
Show answer and solution

Answer: (A) 1 : 1

Solutionr = mv/qB = √(2mK)/qB. Alpha has 4 times the mass (√4 = 2) and twice the charge. The factors cancel, so r_p : r_α = 1 : 1.
Common trapUsing r ∝ m/q, which is true only for equal speeds, not equal kinetic energies.
Current ElectricityMedium

Q8. Two identical cells, each of emf 2 V and internal resistance 1 Ω, are connected in parallel across an external resistor of 1.5 Ω. What current flows through the resistor?

  1. 0.8 A
  2. 1 A
  3. 1.6 A
  4. 2 A
Show answer and solution

Answer: (B) 1 A

SolutionIdentical cells in parallel act as one cell of emf 2 V with internal resistance 1/2 = 0.5 Ω. Current = 2/(1.5 + 0.5) = 1 A.
Common trapAdding the emfs (4 V) as if the cells were in series.
Moving Charges & MagnetismMedium

Q9. A galvanometer of resistance 50 Ω gives full-scale deflection at 1 mA. What shunt resistance converts it into an ammeter of range 0–1 A?

  1. ≈ 0.05 Ω
  2. 0.5 Ω
  3. 5 Ω
  4. ≈ 950 Ω
Show answer and solution

Answer: (A) ≈ 0.05 Ω

SolutionThe shunt carries the remaining current: S = I_g G/(I − I_g) = (0.001 × 50)/0.999 ≈ 0.05 Ω, connected in parallel.
Common trapAbout 950 Ω in series converts it into a voltmeter, not an ammeter.
ElectrostaticsMedium

Q10. A thin spherical shell of radius R carries charge Q spread uniformly on its surface. What is the electric potential at its centre? (k = 1/4πε₀)

  1. 0
  2. kQ/2R
  3. kQ/R
  4. 3kQ/2R
Show answer and solution

Answer: (C) kQ/R

SolutionThe field inside a charged shell is zero, so the potential is the same everywhere inside as on the surface: kQ/R.
Common trapAnswering 0 because the field is zero. Zero field means constant potential, not zero potential. 3kQ/2R is for the centre of a solid charged sphere.
CapacitorsEasy

Q11. Capacitors of 2 μF and 3 μF are connected in series across a 10 V battery. What charge is stored on each?

  1. 6 μC
  2. 12 μC
  3. 20 μC
  4. 50 μC
Show answer and solution

Answer: (B) 12 μC

SolutionSeries capacitance = (2 × 3)/(2 + 3) = 1.2 μF. Charge Q = CV = 1.2 × 10 = 12 μC, the same on both.
Common trap50 μC adds the capacitances as if they were in parallel.
Electromagnetic InductionEasy

Q12. How much energy is stored in an inductor of 2 H carrying a steady current of 3 A?

  1. 3 J
  2. 6 J
  3. 9 J
  4. 18 J
Show answer and solution

Answer: (C) 9 J

SolutionEnergy U = ½LI² = ½ × 2 × 3² = 9 J.
Common trapLeaving out the ½ gives 18 J.
Current ElectricityMediumNumerical

Q13. In a Wheatstone bridge, all four arms are 10 Ω each and a galvanometer of 5 Ω connects the middle points. What is the equivalent resistance (in Ω) between the two input corners?

Numerical value question: type the answer, no options.

Show answer and solution

Answer: 10

SolutionWith equal arms the bridge is balanced, so no current flows through the galvanometer and it can be removed. Two 20 Ω paths in parallel give 10 Ω.
Common trapIncluding the galvanometer's 5 Ω in the calculation. In a balanced bridge it carries no current.

Frequently asked questions

Which Electricity & Magnetism chapters are covered here?

This page covers Alternating Current, Capacitors, Current Electricity, Electromagnetic Induction, Electrostatics, Magnetic Effects of Current, Moving Charges & Magnetism. Questions follow the JEE Main pattern, with multiple-choice and numerical value types.

Are these previous year JEE questions?

No. These are original JEE Main-level practice questions written on the latest pattern, with solutions and the common trap for each.