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Kerala Plus Two Physics (DHSE) study materials
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Kerala Plus Two Physics (DHSE) practice questions

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Question 1Electrostatics

Two point charges of +4 μC and −4 μC are separated by 0.20 m in air. Taking 1/(4πε₀) = 9 × 10⁹ N m² C⁻², what is the magnitude of the force between them?

  1. 1.8 N
  2. 7.2 N
  3. 3.6 N Correct answer.
  4. 0.36 N

Explanation

Coulomb's law gives F = k|q₁q₂|/r². Substituting k = 9 × 10⁹, |q₁q₂| = 16 × 10⁻¹² C² and r² = 0.04 m² gives F = 9 × 10⁹ × 16 × 10⁻¹² / 0.04 = 3.6 N. The opposite signs only fix the direction (attraction), not the magnitude.

Question 2Electrostatics

The electric field at a point on the axial line of an electric dipole of moment p, at distance r from the centre where r ≫ a, is best described as

  1. directed opposite to p and proportional to 1/r²
  2. directed opposite to p and proportional to 1/r³
  3. directed along p and proportional to 1/r³ Correct answer.
  4. directed along p and proportional to 1/r²

Explanation

On the axial line the two dipole fields add, and for r ≫ a the result is E = 2kp/r³, pointing in the direction of the dipole moment. The 1/r³ fall-off is characteristic of a dipole, distinguishing it from the 1/r² field of a single point charge.

Question 3Electrostatics

A hollow conducting sphere carries a net charge. Which statement about the electric field inside the cavity (with no charge inside it) is correct?

  1. The field points radially outward with magnitude kQ/r².
  2. The field is zero only at the centre and grows towards the inner surface.
  3. The field is zero everywhere inside the cavity. Correct answer.
  4. The field is uniform and equal to σ/ε₀.

Explanation

For a conductor in electrostatic equilibrium the field inside the material is zero, and applying Gauss's law to a surface enclosing the empty cavity shows no enclosed charge and hence no flux. Therefore the field vanishes throughout the cavity, and any net charge resides on the outer surface.

Question 4Electrostatics

The work done in moving a charge of 2 μC between two points whose potentials differ by 50 V is

  1. 4 × 10⁻⁴ J
  2. 1 × 10⁻⁴ J Correct answer.
  3. 1 × 10⁻⁶ J
  4. 2.5 × 10⁻⁵ J

Explanation

The work done by the external agent equals qΔV for a slow displacement. Here W = 2 × 10⁻⁶ C × 50 V = 1 × 10⁻⁴ J, independent of the path taken between the two points because the electrostatic field is conservative.

Question 5Electrostatics

A parallel plate capacitor is charged and then disconnected from the battery. If the plate separation is doubled, the energy stored in the capacitor

  1. doubles Correct answer.
  2. becomes four times
  3. halves
  4. remains unchanged

Explanation

With the battery removed the charge Q is fixed, so the energy is U = Q²/2C. Doubling the separation halves the capacitance (C = ε₀A/d), so U doubles; the extra energy comes from the mechanical work done in pulling the plates apart against their attraction.

Question 6Electrostatics

Three charges +q, +q and −q are placed at the vertices A, B and C of an equilateral triangle of side a. The magnitude of the force on the charge at C is

  1. √3 kq²/a² Correct answer.
  2. kq²/a²
  3. 2kq²/a²
  4. kq²/(2a²)

Explanation

Each charge at A and B exerts a force kq²/a² on the charge at C, one attractive along CA and one along CB, with 60° between the two directions. Adding these equal vectors gives a resultant of 2(kq²/a²)cos30° = √3 kq²/a².

Question 7Electrostatics

An electric dipole is placed in a uniform external field with its moment making an angle θ with the field. In which orientation is the dipole in unstable equilibrium?

  1. θ = 90°
  2. θ = 45°
  3. θ = 0°
  4. θ = 180° Correct answer.

Explanation

The torque on a dipole is τ = pE sinθ, which vanishes at θ = 0° and θ = 180°. At θ = 180° the potential energy U = −pE cosθ is maximum, so any small displacement produces a torque that rotates the dipole further away: unstable equilibrium.

Question 8Electrostatics

A charge Q is placed at the centre of a cube of side a. The electric flux through one face of the cube is

  1. Q/(6ε₀) Correct answer.
  2. Q/(4ε₀)
  3. Q/(8ε₀)
  4. Q/ε₀

Explanation

Gauss's law gives total flux Q/ε₀ through the closed surface. By symmetry the flux is shared equally among the six faces of the cube, so each face carries Q/(6ε₀).

Question 9Electrostatics

Two identical capacitors, each of capacitance C, are first connected in series and then in parallel across the same battery. The ratio of energy stored in the parallel combination to that in the series combination is

  1. 4 : 1 Correct answer.
  2. 2 : 1
  3. 1 : 4
  4. 1 : 2

Explanation

Series gives C/2 and parallel gives 2C. For the same voltage, energy is ½CV², so the ratio is 2C ÷ (C/2) = 4. The parallel arrangement stores four times the energy.

Question 10Electrostatics

The electric potential due to a short dipole at a point on its equatorial line, at distance r ≫ a, is

  1. zero Correct answer.
  2. kp/r
  3. 2kp/r²
  4. kp/r²

Explanation

On the equatorial line the point is equidistant from +q and −q, so the potentials kq/r and −kq/r cancel exactly. The potential is therefore zero everywhere on the equatorial plane, even though the electric field there is non-zero and points opposite to p.

Question 11Electrostatics

A charge q is placed at the centre of a spherical Gaussian surface. If the radius of the surface is doubled while the charge stays at the centre, which quantity remains unchanged?

  1. The surface area of the sphere
  2. The field energy density at the surface
  3. The total electric flux through the surface Correct answer.
  4. The electric field at the surface

Explanation

Gauss's law states that the flux through a closed surface equals the enclosed charge divided by ε₀, so it depends only on q and is unaffected by the size of the surface. The field strength, area and energy density all change with radius.

Question 12Electrostatics

An isolated charged conductor of irregular shape is in electrostatic equilibrium. Which statement is correct?

  1. The surface charge density is greatest where the curvature is greatest. Correct answer.
  2. The surface charge density is greatest where the curvature is least.
  3. The surface charge density is uniform over the whole surface.
  4. The surface charge density is greatest on the flattest portion.

Explanation

In equilibrium the conductor surface is an equipotential, and the field just outside is σ/ε₀. To keep the potential the same, charge concentrates at sharp points where the curvature is large, giving the highest surface charge density there. This is why corona discharge occurs at pointed ends.

Question 13Electrostatics

A parallel plate capacitor has capacitance C₀ in air. A dielectric slab of constant K = 3 fills the space completely. If the capacitor is kept connected to the same battery, the stored energy becomes

  1. 3 C₀V²/2 Correct answer.
  2. C₀V²/6
  3. C₀V²/2
  4. 3 C₀V²

Explanation

With the battery connected the voltage V is fixed, and inserting the dielectric raises the capacitance to KC₀ = 3C₀. The energy ½CV² therefore becomes ½ × 3C₀V² = 3C₀V²/2, with the extra energy supplied by the battery as it pushes more charge onto the plates.

Question 14Electrostatics

Four charges, each +q, are placed at the corners of a square of side a. The work needed to bring a fifth charge +Q from infinity to the centre of the square is

  1. 4√2 kqQ/a Correct answer.
  2. zero
  3. 4kqQ/a
  4. 2√2 kqQ/a

Explanation

The centre is at distance a/√2 from each corner, so each corner charge contributes potential √2 kq/a. The total potential is 4√2 kq/a, and the work done is Q times this value, giving 4√2 kqQ/a.

Question 15Electrostatics

A capacitor of capacitance C is charged to potential V and then connected across an identical uncharged capacitor. The energy lost as heat in the connecting wires is

  1. CV²/8
  2. CV²/2
  3. zero
  4. CV²/4 Correct answer.

Explanation

The initial energy is ½CV². After connection the charge CV is shared equally, so each capacitor holds CV/2 at a common potential V/2, giving a final total energy of 2 × ½C(V/2)² = CV²/4. The loss is therefore ½CV² − ¼CV² = CV²/4, appearing as heat and radiation in the wires.

Question 16Electrostatics

A parallel plate capacitor with air between its plates has capacitance C. A slab of dielectric constant K = 4 and thickness equal to half the plate separation is placed between the plates, covering the whole area. The new capacitance is

  1. 5C/4
  2. 4C
  3. 2C
  4. 8C/5 Correct answer.

Explanation

The arrangement is equivalent to two capacitors in series: an air gap of thickness d/2 giving capacitance 2C, and a dielectric layer of thickness d/2 giving 2KC = 8C. For series combination the reciprocal adds: 1/C' = 1/(2C) + 1/(8C) = 5/(8C), so C' = 8C/5. A partially filled capacitor therefore never exceeds K times the air value, and here the air region in series limits the increase.

Question 17Current Electricity

A copper wire of length L and uniform cross-sectional area A has resistance R. It is stretched uniformly so that its new length is 2L. Assuming the volume and resistivity remain unchanged, what is the resistance of the stretched wire?

  1. 4R Correct answer.
  2. 2R
  3. 8R
  4. R/2

Explanation

Stretching at constant volume doubles the length and halves the cross-sectional area, since A' = V/(2L) = A/2. Resistance is R = ρL/A, so the new value is ρ(2L)/(A/2) = 4ρL/A = 4R. The resistance therefore rises by the square of the length-stretch factor.

Question 18Current Electricity

The drift velocity of free electrons in a metallic conductor carrying a steady current is best described as:

  1. The root-mean-square speed of the electrons due to thermal agitation alone
  2. The speed of the electric field signal propagating through the conductor
  3. The average velocity acquired by electrons along the direction of the applied field, superimposed on their random thermal motion Correct answer.
  4. The speed at which an individual electron travels from one end of the wire to the other

Explanation

In a metal, electrons move randomly with high thermal speeds, and the applied field adds a small systematic drift opposite to the field. Drift velocity is the average of this net directional motion, of order a fraction of a millimetre per second. It is not the speed of any single electron nor the speed of the electrical signal.

Question 19Current Electricity

Two cells of emf 2.0 V and 1.5 V with internal resistances 0.5 Ω and 0.5 Ω respectively are connected in parallel across a 4.0 Ω resistor. Using the parallel-cell formula, the current through the resistor is approximately:

  1. 0.70 A
  2. 0.82 A Correct answer.
  3. 0.88 A
  4. 0.44 A

Explanation

For two cells in parallel, the equivalent emf is (E1r2 + E2r1)/(r1 + r2) = (2.0×0.5 + 1.5×0.5)/(1.0) = 1.75 V, and the equivalent internal resistance is (r1r2)/(r1 + r2) = 0.25 Ω. The total resistance is 4.25 Ω, giving I = 1.75/4.25 ≈ 0.82 A.

Question 20Current Electricity

Kirchhoff's junction rule is a direct consequence of which conservation principle?

  1. Conservation of electric charge Correct answer.
  2. Conservation of magnetic flux
  3. Conservation of momentum
  4. Conservation of energy

Explanation

The junction rule states that the algebraic sum of currents meeting at a node is zero, meaning charge does not accumulate at the junction. This follows from conservation of charge in a steady state. Conservation of energy underlies the loop rule instead.

Question 21Current Electricity

A wire of resistance R is bent into a closed circle and two diametrically opposite points are connected to a battery of negligible internal resistance. What is the equivalent resistance between these two points?

  1. R/2
  2. 2R
  3. R/4 Correct answer.
  4. R

Explanation

The two diametrically opposite points divide the ring into two semicircular halves, each of resistance R/2. These two halves are in parallel between the same pair of terminals. Hence the equivalent resistance is (R/2 × R/2)/(R/2 + R/2) = R/4.

Question 22Current Electricity

The terminal potential difference of a cell of emf E and internal resistance r, when it delivers current I to an external circuit, is given by:

  1. E/r
  2. E − Ir Correct answer.
  3. Ir − E
  4. E + Ir

Explanation

When a cell drives current through an external resistor, part of its emf is dropped across the internal resistance as Ir. The terminal voltage available to the external circuit is therefore E − Ir. It is always less than the emf while the cell is discharging.

Question 23Current Electricity

A potentiometer wire of length 10 m and resistance 20 Ω is connected in series with a 2 V driver cell and a 380 Ω resistor. What is the potential gradient along the wire?

  1. 1.0 V/m
  2. 0.05 V/m
  3. 0.1 V/m Correct answer.
  4. 0.2 V/m

Explanation

The total circuit resistance is 380 + 20 = 400 Ω, so the current is 2/400 = 0.005 A. The potential drop across the 20 Ω wire is 0.005 × 20 = 0.1 V. Dividing by the 10 m length gives a potential gradient of 0.01 V per metre... wait, 0.1 V over 10 m is 0.01 V/m.

Question 24Current Electricity

In a metre bridge experiment, the balance point is obtained at 40 cm from the left end when an unknown resistance X is in the left gap and a 6 Ω standard resistor is in the right gap. The value of X is:

  1. 4 Ω Correct answer.
  2. 6 Ω
  3. 2.4 Ω
  4. 9 Ω

Explanation

For a metre bridge at balance, X/R = l/(100 − l), where l is the balancing length from the end where X is connected. Here X/6 = 40/60 = 2/3, so X = 6 × 2/3 = 4 Ω.

Question 25Current Electricity

The resistivity of a metal conductor increases with temperature primarily because:

  1. Increased lattice vibrations cause more frequent collisions of the conduction electrons Correct answer.
  2. The length of the conductor increases due to thermal expansion
  3. The number density of free electrons decreases sharply with temperature
  4. The charge on each electron decreases as temperature rises

Explanation

In a metal, the free-electron density is essentially independent of temperature. Heating increases the amplitude of lattice vibrations, so electrons collide more often and the relaxation time falls, raising resistivity. The small thermal-expansion change in dimensions is not the dominant cause.

Question 26Current Electricity

Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel. The equivalent resistance of the combination is:

  1. 11 Ω
  2. 0.5 Ω
  3. 3.67 Ω
  4. 1 Ω Correct answer.

Explanation

For parallel resistors, 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R = 1 Ω. The equivalent resistance is always smaller than the smallest individual resistance in a parallel network.

Question 27Current Electricity

A cell of emf 6 V and internal resistance 1 Ω is connected to a variable external resistance R. At what value of R is the power delivered to the external resistance maximum?

  1. 6 Ω
  2. 0 Ω
  3. 36 Ω
  4. 1 Ω Correct answer.

Explanation

The power in the external resistor is P = E²R/(R + r)². Differentiating with respect to R and setting the derivative to zero gives R = r, the condition for maximum power transfer. Here r = 1 Ω, so maximum power occurs at R = 1 Ω.

Question 28Current Electricity

A galvanometer of resistance 50 Ω gives full-scale deflection for a current of 1 mA. To convert it into a voltmeter reading up to 10 V, the resistance to be connected in series is:

  1. 10 Ω
  2. 9950 Ω Correct answer.
  3. 50 Ω
  4. 10000 Ω

Explanation

For a voltmeter, a high resistance R is placed in series so that the full-scale current flows when 10 V is applied. Thus 10 = Ig(R + G) = 0.001(R + 50), giving R + 50 = 10000 Ω and R = 9950 Ω.

Question 29Current Electricity

In a Wheatstone bridge, the four arms are P, Q, R and S arranged as usual, with a galvanometer between the junctions. The bridge is balanced. If the cell and the galvanometer are now interchanged, the bridge:

  1. Shows a deflection because the current directions reverse everywhere
  2. Becomes unbalanced unless P = Q
  3. Becomes unbalanced unless all four arms are equal
  4. Remains balanced, since the balance condition P/Q = R/S is symmetric under this interchange Correct answer.

Explanation

The balance condition of a Wheatstone bridge is P/Q = R/S, which is unchanged when the battery and galvanometer branches are swapped. This symmetry means a balanced bridge stays balanced after the interchange. It is a standard property exploited in bridge measurements.

Question 30Current Electricity

The current density J in a conductor is related to the electric field E and conductivity σ by:

  1. J = E/σ
  2. J = σE²
  3. J = σ/E
  4. J = σE Correct answer.

Explanation

This is the microscopic form of Ohm's law. Current density is proportional to the applied field, with conductivity as the constant of proportionality. In terms of resistivity ρ, the same relation is written J = E/ρ.

Question 31Magnetic Effects of Current and Magnetism

A long straight wire carries a steady current I. According to Ampere's circuital law, the magnitude of the magnetic field at a perpendicular distance r from the wire in free space is given by which expression?

  1. B = μ₀I/(2πr) Correct answer.
  2. B = μ₀I/(4πr)
  3. B = μ₀I/(2r²)
  4. B = μ₀I r/(2π)

Explanation

Ampere's circuital law applied to a circular loop of radius r around the wire gives B(2πr) = μ₀I, so B = μ₀I/(2πr). The field falls off as 1/r and is tangential to circles centred on the wire.

Question 32Magnetic Effects of Current and Magnetism

A charged particle enters a uniform magnetic field with its velocity perpendicular to the field. The path followed by the particle is

  1. a circle whose radius is independent of the field strength
  2. a straight line along the field direction
  3. a parabola opening along the field direction
  4. a circle whose radius decreases as the field strength increases Correct answer.

Explanation

The magnetic force qvB acts perpendicular to velocity, providing centripetal force mv²/r, so r = mv/(qB). Hence the radius is inversely proportional to B and decreases when the field is strengthened.

Question 33Magnetic Effects of Current and Magnetism

The magnetic field at the centre of a circular coil of N turns, radius R, carrying current I is

  1. B = μ₀NI/(2R) Correct answer.
  2. B = μ₀I/(2πR)
  3. B = μ₀NI/(2πR)
  4. B = μ₀NI/(4R)

Explanation

Each turn contributes μ₀I/(2R) at the centre, and N turns add in the same direction, giving B = μ₀NI/(2R). The field is perpendicular to the plane of the coil, directed by the right-hand rule.

Question 34Magnetic Effects of Current and Magnetism

The SI unit of magnetic dipole moment is

  1. A m² Correct answer.
  2. N m A⁻¹
  3. T m²
  4. A m⁻¹

Explanation

Magnetic moment m = NIA, so its unit is ampere times square metre, i.e. A m². Equivalently it is joule per tesla, since torque τ = mB sinθ.

Question 35Magnetic Effects of Current and Magnetism

The force per unit length between two long parallel straight wires carrying currents I₁ and I₂ separated by distance d in free space is

  1. μ₀I₁I₂/(4πd²)
  2. μ₀I₁I₂ d/(2π)
  3. μ₀I₁I₂/(2πd) Correct answer.
  4. μ₀(I₁ + I₂)/(2πd)

Explanation

The field of one wire at the other is μ₀I₁/(2πd), and the force on unit length of the second wire carrying I₂ is B I₂, giving μ₀I₁I₂/(2πd). The force is attractive for parallel currents and repulsive for antiparallel currents.

Question 36Magnetic Effects of Current and Magnetism

A galvanometer of resistance 50 Ω gives a full-scale deflection for a current of 2 mA. To convert it into an ammeter of range 0 to 2 A, the shunt resistance required is nearly

  1. 25.0 Ω
  2. 0.05 Ω Correct answer.
  3. 0.50 Ω
  4. 50.0 Ω

Explanation

The shunt carries the excess current 2 A − 0.002 A ≈ 2 A while the galvanometer carries 2 mA across the same potential difference. Thus S = (I_g R_g)/(I − I_g) ≈ (0.002 × 50)/2 = 0.05 Ω. A small shunt in parallel diverts most of the current.

Question 37Magnetic Effects of Current and Magnetism

A galvanometer of resistance 100 Ω shows full-scale deflection for 1 mA. To convert it into a voltmeter of range 0 to 10 V, the series resistance that must be added is

  1. 9 000 Ω
  2. 9 900 Ω Correct answer.
  3. 10 000 Ω
  4. 100 Ω

Explanation

For full-scale deflection, the total resistance must be R_total = V/I_g = 10/0.001 = 10 000 Ω. Subtracting the galvanometer resistance gives the series resistance R = 10 000 − 100 = 9 900 Ω.

Question 38Magnetic Effects of Current and Magnetism

Two long parallel wires carry currents in the same direction. The nature of the force between them and the magnetic field at a point midway between them are respectively

  1. attractive and zero Correct answer.
  2. repulsive and maximum
  3. repulsive and zero
  4. attractive and maximum

Explanation

Parallel currents in the same direction attract each other. At the midpoint the fields due to the two wires are equal in magnitude but opposite in direction, so they cancel and the resultant field is zero.

Question 39Magnetic Effects of Current and Magnetism

A circular coil of 100 turns and radius 0.1 m carries a current of 0.5 A. Its magnetic dipole moment is nearly

  1. 0.157 A m²
  2. 15.7 A m²
  3. 0.0157 A m²
  4. 1.57 A m² Correct answer.

Explanation

Magnetic moment m = NIA, with area A = πr² = π(0.1)² ≈ 0.0314 m². Thus m = 100 × 0.5 × 0.0314 ≈ 1.57 A m².

Question 40Magnetic Effects of Current and Magnetism

A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is

  1. zero Correct answer.
  2. directed perpendicular to both v and B
  3. maximum, equal to qvB
  4. qvB/2

Explanation

The magnetic force is F = qvB sinθ, where θ is the angle between velocity and field. When the particle moves parallel to B, θ = 0 and sinθ = 0, so the force vanishes and the particle continues in a straight line.

Question 41Magnetic Effects of Current and Magnetism

A magnetic dipole of moment m is placed in a uniform magnetic field B, with m making an angle θ with B. The potential energy of the dipole, taking zero at θ = 90°, is

  1. mB sinθ
  2. −mB cosθ Correct answer.
  3. mB cosθ
  4. −mB sinθ

Explanation

The torque is τ = mB sinθ, and the work done in rotating the dipole from 90° to θ gives U = −mB cosθ. This is minimum (−mB) when m is aligned with B and maximum (+mB) when antiparallel.

Question 42Magnetic Effects of Current and Magnetism

A bar magnet is dropped through a horizontal copper ring held with its plane horizontal. As the magnet approaches the ring from above, the induced current in the ring, viewed from above, and the force on the magnet are

  1. anticlockwise, with the ring attracting the magnet
  2. anticlockwise, with the ring repelling the magnet Correct answer.
  3. clockwise, with the ring repelling the magnet
  4. clockwise, with the ring attracting the magnet

Explanation

As the north pole approaches from above, the flux downward through the ring increases, so Lenz's law makes the ring develop an upward field, i.e. its upper face acts as a north pole. Viewed from above this requires an anticlockwise current, and the like poles repel the falling magnet.

Question 43Magnetic Effects of Current and Magnetism

The magnetic field on the axis of a circular coil of radius R carrying current I, at a distance x from its centre along the axis, is proportional to

  1. (R² + x²)^(3/2) in the denominator Correct answer.
  2. (R² + x²) in the denominator
  3. (R² − x²)^(3/2) in the denominator
  4. (R² + x²)^(1/2) in the denominator

Explanation

The axial field is B = μ₀NIR²/[2(R² + x²)^(3/2)]. The 3/2 power arises because the axial field component of each element falls off with the cube of the distance from the element to the point.

Question 44Magnetic Effects of Current and Magnetism

A current-carrying coil is suspended in a radial magnetic field produced by concave pole pieces, and a soft-iron core is placed inside it. The main purpose of this arrangement in a moving-coil galvanometer is to

  1. make the field non-uniform so that the torque depends on the angle of deflection
  2. eliminate the need for a restoring spring in the instrument
  3. increase the resistance of the coil so that the current is reduced
  4. keep the plane of the coil always parallel to the field, giving a uniform deflecting torque Correct answer.

Explanation

With concave poles and a soft-iron core the field is radial, so the plane of the coil stays parallel to the field at every deflection. The torque NBiA is then constant for a given current, giving a linear scale, and the core strengthens the field.

Question 45Magnetic Effects of Current and Magnetism

Two infinitely long straight parallel wires carry equal currents I in opposite directions and are separated by distance 2a. The magnitude of the magnetic field at a point midway between them is

  1. zero
  2. μ₀I/(2πa)
  3. μ₀I/(4πa)
  4. μ₀I/(πa) Correct answer.

Explanation

At the midpoint each wire is at distance a, giving a field μ₀I/(2πa). Because the currents are opposite, the two fields at the midpoint point in the same direction and add, giving μ₀I/(πa).

Question 46Magnetic Effects of Current and Magnetism

A charged particle enters a region containing both a uniform electric field and a uniform magnetic field, with velocity, E and B mutually perpendicular. For a particular speed the particle passes undeviated. This speed is

  1. E/B²
  2. EB
  3. E/B Correct answer.
  4. B/E

Explanation

Undeviated motion requires the electric force qE and magnetic force qvB to be equal and opposite, so qE = qvB and v = E/B. This is the principle of the velocity selector used in devices such as the cyclotron source region.

Question 47Electromagnetic Induction and Alternating Currents

A straight conductor of length 0.5 m moves with a velocity of 4 m/s at right angles to a uniform magnetic field of 0.2 T. What is the magnitude of the emf induced across its ends?

  1. 0.8 V
  2. 0.4 V Correct answer.
  3. 4 V
  4. 0.04 V

Explanation

For motional emf, the induced emf equals B times l times v when the velocity, field and conductor are mutually perpendicular. Here 0.2 × 0.5 × 4 gives 0.4 V. The direction would follow Fleming's right-hand rule, but only the magnitude is asked.

Question 48Electromagnetic Induction and Alternating Currents

Lenz's law is a direct consequence of which conservation principle applied to electromagnetic induction?

  1. Conservation of magnetic flux
  2. Conservation of energy Correct answer.
  3. Conservation of momentum
  4. Conservation of charge

Explanation

The induced current always opposes the change producing it, so mechanical work must be done against the magnetic force to generate electrical energy. If the induced current aided the change, energy would be created without any input. Hence Lenz's law expresses conservation of energy.

Question 49Electromagnetic Induction and Alternating Currents

The self-inductance of a coil is 2 H and the current through it changes uniformly from 0 to 5 A in 0.1 s. What is the magnitude of the self-induced emf?

  1. 1 V
  2. 0.1 V
  3. 100 V Correct answer.
  4. 10 V

Explanation

The self-induced emf magnitude is L times the rate of change of current. The rate is 5 A divided by 0.1 s, that is 50 A/s, and multiplying by L = 2 H gives 100 V. The negative sign in the full formula only indicates opposition to the change.

Question 50Electromagnetic Induction and Alternating Currents

In a purely capacitive AC circuit, the current with respect to the applied voltage behaves in which way?

  1. It leads the voltage by 90 degrees. Correct answer.
  2. It leads the voltage by 45 degrees.
  3. It lags the voltage by 90 degrees.
  4. It is in phase with the voltage.

Explanation

In a capacitor the current is proportional to the rate of change of charge, and the derivative of a sine is a cosine, giving a quarter-cycle advance. So the current reaches its peak a quarter period before the voltage. This is summarised as current leading voltage by 90 degrees.