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Assam CEE Electromagnetism — practice questions — page 2

Explore 89 free Electromagnetism questions in Assam CEE Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Electromagnetism questions and exercises with answers

Showing questions 31–60 of 89.

For a parallel plate capacitor with plate area A and separation d, the capacitance in vacuum:Read the answer and worked solution →Capacitance changes when DIELECTRIC of relative permittivity K is inserted:Read the answer and worked solution →Energy stored in a capacitor with charge Q and capacitance C:Read the answer and worked solution →Two charges +q and +q are placed at corners of an equilateral triangle. The third corner has potential:Read the answer and worked solution →For a spherical conductor of radius R carrying charge Q, the potential AT THE SURFACE equals:Read the answer and worked solution →Equipotential surfaces around a point charge are:Read the answer and worked solution →Two capacitors 2 µF and 4 µF connected in SERIES across 12 V. Charge on each:Read the answer and worked solution →3 capacitors of 6 µF each in PARALLEL: total capacitance:Read the answer and worked solution →Two charged spheres each of radius R, charge Q, separated by distance d >> R. Force F between them is:Read the answer and worked solution →Potential at center of square (side a) with charges +q, -q, +q, -q at corners (alternating):Read the answer and worked solution →Electric field inside a spherical cavity in a uniformly charged conductor is:Read the answer and worked solution →Find work done to bring a +2 µC charge from infinity to a point where V = 50 V:Read the answer and worked solution →A charged particle q is held at rest in vertical equilibrium in a uniform electric field E pointing upward. Charge-to-mass ratio q/m:Read the answer and worked solution →In a capacitor circuit, when a dielectric is inserted while voltage is HELD CONSTANT by battery, the CHARGE on capacitor:Read the answer and worked solution →Magnetic field is measured in:Read the answer and worked solution →Biot Savart law gives magnetic field due to a current element. It states dB is proportional to:Read the answer and worked solution →The magnetic force on a charge q moving with velocity v in field B:Read the answer and worked solution →A current-carrying wire of length L in magnetic field B experiences force:Read the answer and worked solution →Ampere's law states the line integral of B around a closed loop equals:Read the answer and worked solution →A magnetic dipole moment of a current loop is:Read the answer and worked solution →Magnetic field at center of a circular loop of radius R carrying current I:Read the answer and worked solution →Magnetic field inside a long solenoid (n turns per unit length, current I):Read the answer and worked solution →Magnetic field at distance r from a long straight current-carrying wire:Read the answer and worked solution →A charged particle of charge q, mass m enters magnetic field B perpendicular to velocity v. It follows a:Read the answer and worked solution →Torque on a current loop (moment m) in uniform field B:Read the answer and worked solution →In a velocity selector, electric field E and magnetic field B are perpendicular. Velocity of particles passing undeflected:Read the answer and worked solution →Magnetic field on the axis of a circular loop at distance x from center:Read the answer and worked solution →Magnetic flux through a surface is:Read the answer and worked solution →Cyclotron frequency for a charged particle in magnetic field B:Read the answer and worked solution →A solenoid of 200 turns, length 0.5 m, carries 2 A. Find magnetic field inside (μ₀ = 4π × 10⁻⁷):Read the answer and worked solution →

Worked practice questions

For a parallel plate capacitor with plate area A and separation d, the capacitance in vacuum:

  1. C = ε₀A/d
  2. C = ε₀d/A
  3. C = A/(ε₀d)
  4. C = dε₀A
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Correct answer: A. C = ε₀A/d

C = ε₀A/d. Larger area or smaller separation → more capacitance. Insertion of dielectric multiplies capacitance by relative permittivity ε_r.

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Capacitance changes when DIELECTRIC of relative permittivity K is inserted:

  1. Decreases by K
  2. Stays same
  3. Increases by factor K (C → KC)
  4. Becomes zero
Show answer and solution

Correct answer: C. Increases by factor K (C → KC)

**With dielectric:** C = Kε₀A/d. Capacitance becomes K times the vacuum value. Energy stored at same voltage also increases (since U = ½CV²).

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Energy stored in a capacitor with charge Q and capacitance C:

  1. Q²/C
  2. Q²/(2C)
  3. CQ²
  4. 2CQ²
Show answer and solution

Correct answer: B. Q²/(2C)

We have U = Q²/(2C). That is U = (1/2)CV². So U = (1/2)QV. Three equivalent forms — useful depending on which two variables you know.

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