I85PREP
⌘ K
RBSE SS Physics study materials
RBSE SS Physics study resources

RBSE SS Physics practice questions

One question at a time. Build your RBSE SS Physics understanding.

First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Electrostatics

Two point charges of +2 μC and +8 μC are separated by 0.2 m in vacuum. Using Coulomb's constant k = 9 × 10^9 N·m²/C², what is the magnitude of the electrostatic force between them?

  1. 1.8 N
  2. 7.2 N
  3. 14.4 N
  4. 3.6 N Correct answer.

Explanation

Coulomb's law gives F = k·q₁·q₂/r². Substituting (9 × 10^9 × 2 × 10^-6 × 8 × 10^-6)/(0.2)² = (0.144)/(0.04) = 3.6 N. The force is repulsive because both charges are positive.

Question 2Electrostatics

A test charge is placed at a point where the electric field due to a fixed source charge is directed toward the east. If the test charge is negative, in which direction does the electrostatic force on it act?

  1. The force is zero
  2. Toward the north
  3. Toward the east
  4. Toward the west Correct answer.

Explanation

The force on a charge is F = qE, so a negative q reverses the direction of the field. Since the field points east, the force on a negative test charge points west.

Question 3Electrostatics

The electric flux through a closed surface enclosing no net charge is zero. Which statement best explains this result?

  1. The permittivity of free space is zero
  2. The field inside the surface is always zero
  3. Every field line that enters the surface also leaves it Correct answer.
  4. The surface must be spherical for flux to be defined

Explanation

Gauss's law states that net flux through a closed surface equals enclosed charge divided by ε₀. With no enclosed charge, every field line entering the volume must exit it, so inward and outward contributions cancel.

Question 4Electrostatics

Two identical small conducting spheres carry charges +6 μC and −2 μC. They are touched together and then separated to their original distance. What are the final charges on the spheres?

  1. +4 μC and −4 μC
  2. +6 μC and −2 μC
  3. +2 μC and +2 μC Correct answer.
  4. +3 μC and +1 μC

Explanation

When identical conductors touch, total charge is shared equally. The total is (+6) + (−2) = +4 μC, so each sphere ends with +2 μC.

Question 5Electrostatics

The work done by the external agent in moving a charge of 5 μC from infinity to a point at potential 200 V is:

  1. 2.5 × 10^-8 J
  2. 1 × 10^-6 J
  3. 1 × 10^-3 J Correct answer.
  4. 4 × 10^-4 J

Explanation

Work done equals q times the potential difference, W = qV. Here W = 5 × 10^-6 C × 200 V = 1 × 10^-3 J, since potential at infinity is taken as zero.

Question 6Electrostatics

The electric potential at a distance r from an isolated point charge Q is V = kQ/r. How does the potential vary with distance compared with the field?

  1. Both potential and field fall as 1/r²
  2. Potential falls as 1/r while the field falls as 1/r² Correct answer.
  3. Both potential and field fall as 1/r
  4. Potential falls as 1/r² while the field falls as 1/r

Explanation

For a point charge, V = kQ/r varies inversely with r, whereas E = kQ/r² varies inversely with r squared. The field is the negative gradient of potential, which accounts for the extra power of r.

Question 7Electrostatics

A parallel-plate capacitor has capacitance C with plate separation d and air between the plates. If the separation is doubled while the plates remain connected to a battery, what happens to the stored charge?

  1. It doubles
  2. It is halved Correct answer.
  3. It stays the same
  4. It becomes four times larger

Explanation

For a parallel-plate capacitor, C = ε₀A/d, so doubling d halves the capacitance. Because the battery keeps V fixed, Q = CV is also halved.

Question 8Electrostatics

A capacitor of 10 μF is charged to 100 V. What energy is stored in it?

  1. 0.05 J Correct answer.
  2. 1.0 J
  3. 0.5 J
  4. 5 × 10^-3 J

Explanation

Energy stored in a capacitor is U = ½CV². Substituting gives ½ × 10 × 10^-6 × (100)² = ½ × 10^-5 × 10^4 = 0.05 J.

Question 9Electrostatics

The electric field just outside the surface of a charged conductor in electrostatic equilibrium is directed:

  1. At 45° to the surface
  2. Randomly with respect to the surface
  3. Perpendicular to the surface Correct answer.
  4. Parallel to the surface

Explanation

In equilibrium, charges on a conductor arrange so that no field exists inside and no tangential field exists at the surface. Any tangential component would drive surface charges to move, so the field just outside must be normal to the surface.

Question 10Electrostatics

Two capacitors of 4 μF and 12 μF are connected in series across a 16 V supply. What is the equivalent capacitance?

  1. 16 μF
  2. 3 μF Correct answer.
  3. 8 μF
  4. 48 μF

Explanation

For series capacitors, 1/C = 1/C₁ + 1/C₂ = 1/4 + 1/12 = 4/12, so C = 3 μF. The equivalent capacitance in series is always smaller than the smallest individual capacitance.

Question 11Electrostatics

A dipole of moment p is placed in a uniform electric field E so that p is parallel to E. Which statement describes the torque and potential energy in this orientation?

  1. Torque is zero and potential energy is maximum
  2. Torque is maximum and potential energy is maximum
  3. Torque is maximum and potential energy is minimum
  4. Torque is zero and potential energy is minimum Correct answer.

Explanation

Torque on a dipole is τ = pE sin θ and potential energy is U = −pE cos θ. When p is parallel to E, θ = 0, so sin θ = 0 (no torque) and cos θ = 1, giving the most negative, hence minimum, potential energy.

Question 12Electrostatics

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

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

Explanation

By Gauss's law the total flux through the closed cube is Q/ε₀. The central position gives six identical faces, so each face carries one-sixth of the total flux, Q/(6ε₀).

Question 13Electrostatics

A hollow conducting sphere carries a net charge +Q on its outer surface. A point charge +q is placed inside the cavity without touching the wall. What is the field inside the conductor material?

  1. Directed radially inward
  2. Equal to the field at the cavity centre
  3. Zero Correct answer.
  4. Directed radially outward

Explanation

In electrostatic equilibrium the electric field everywhere inside the conducting material is zero. The inner surface acquires charge −q to cancel the cavity field within the metal, and the outer surface holds +Q + q.

Question 14Electrostatics

A parallel-plate capacitor is charged and then disconnected from the battery. A dielectric slab of constant K is inserted to fill the gap. What happens to the potential difference and the stored energy?

  1. Potential decreases by K and energy stays constant
  2. Both increase by a factor K
  3. Both decrease by a factor K Correct answer.
  4. Potential stays constant and energy decreases by K

Explanation

With the battery disconnected, charge Q is fixed. Inserting the dielectric multiplies capacitance by K, so V = Q/C falls by a factor K and U = Q²/(2C) also falls by K.

Question 15Electrostatics

An electric dipole is placed in a non-uniform electric field. Which combination of effects does it generally experience?

  1. Torque only
  2. Net force only
  3. Both a net torque and a net force Correct answer.
  4. Neither a torque nor a force

Explanation

In a non-uniform field the two ends of the dipole feel unequal forces, giving a net force, and any misalignment of p with E gives a torque. In a uniform field only the torque survives, since the forces cancel.

Question 16Electrostatics

Which factor determines the direction of the electric field at a point due to a positive point charge?

  1. Along a circle around the charge
  2. Radially away from the charge Correct answer.
  3. Depends on the magnitude of the test charge
  4. Radially toward the charge

Explanation

By convention the field direction is the direction of force on a positive test charge. A positive source charge repels such a test charge, so the field points radially outward from it.

Question 17Current Electricity

A copper wire of length L and cross-sectional area A has resistance R. If it is drawn out uniformly so that its new length is 2L while its volume stays constant, what is its new resistance?

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

Explanation

Volume is constant, so doubling the length halves the cross-sectional area. Resistance depends on length divided by area, so both changes act in the same direction: the factor is 2 × 2 = 4.

Question 18Current Electricity

Three resistors of 2 Ω, 3 Ω and 6 Ω are joined in parallel. What is the equivalent resistance of the combination?

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

Explanation

For parallel resistors the reciprocals add: 1/R = 1/2 + 1/3 + 1/6 = 1. The equivalent resistance is therefore 1 Ω, which is smaller than the smallest branch resistor.

Question 19Current Electricity

A cell of emf 12 V and internal resistance 2 Ω is connected to an external resistance of 4 Ω. What is the terminal potential difference across the cell?

  1. 8 V Correct answer.
  2. 6 V
  3. 12 V
  4. 4 V

Explanation

The current is emf divided by total resistance, 12/6 = 2 A. Terminal voltage equals emf minus the internal drop, 12 − (2 × 2) = 8 V, which also equals the voltage across the 4 Ω load.

Question 20Current Electricity

Which statement correctly describes the drift velocity of free electrons in a metallic conductor carrying a steady current?

  1. It equals the speed of light in the metal
  2. It is of the order of a few millimetres per second and is opposite to the conventional current direction Correct answer.
  3. It is of the order of a few millimetres per second and is in the same direction as the conventional current
  4. It increases without limit as the wire gets longer

Explanation

Electrons are negatively charged, so their drift is opposite to the conventional current. Typical drift speeds are only a fraction of a millimetre per second, far below the random thermal speeds.

Question 21Current Electricity

The resistance of a metallic conductor increases with temperature mainly because

  1. the ions of the lattice expand and the wire lengthens only
  2. the number density of free electrons decreases sharply
  3. lattice vibrations increase, so electrons collide more frequently and relaxation time falls Correct answer.
  4. the drift velocity of electrons falls to zero

Explanation

In a metal the carrier density is essentially fixed, so resistivity depends on how often electrons scatter. Higher temperature means stronger lattice vibrations, shorter relaxation time and therefore greater resistance.

Question 22Current Electricity

A wire of resistance 10 Ω is bent into a closed circle. What is the resistance between two points on the circle that are diametrically opposite?

  1. 10 Ω
  2. 20 Ω
  3. 5 Ω
  4. 2.5 Ω Correct answer.

Explanation

The two halves of the loop, each of 5 Ω, form two parallel paths between the diametrically opposite points. Their equivalent resistance is 5/2 = 2.5 Ω.

Question 23Current Electricity

Kirchhoff's junction rule is a direct statement of which conservation law for a steady-current circuit?

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

Explanation

The junction rule states that the total current entering a node equals the total leaving it. Charge cannot accumulate at a steady-state junction, so the rule expresses conservation of charge.

Question 24Current Electricity

A battery of emf 6 V and internal resistance 1 Ω delivers maximum power to an external resistor. What is the value of that external resistance?

  1. 1 Ω Correct answer.
  2. 2 Ω
  3. 6 Ω
  4. 0.5 Ω

Explanation

Power in the load is greatest when the external resistance equals the internal resistance, a condition called impedance matching. Here the internal resistance is 1 Ω, so the load must also be 1 Ω.

Question 25Current Electricity

The current through a 4 Ω resistor is 3 A. At what rate is electrical energy converted into heat in the resistor?

  1. 24 W
  2. 12 W
  3. 48 W
  4. 36 W Correct answer.

Explanation

Joule heating power is I²R, so 3² × 4 = 36 W. Equivalently, the potential difference is 12 V and P = VI = 12 × 3 = 36 W.

Question 26Current Electricity

A student measures the potential difference across a resistor as 5.0 V and the current through it as 0.25 A. What is the resistance, expressed to the appropriate significant figures?

  1. 20.0 Ω
  2. 20 Ω Correct answer.
  3. 1.25 Ω
  4. 0.05 Ω

Explanation

From Ohm's law R = V/I = 5.0/0.25 = 20 Ω. Both readings have two significant figures, so the result is quoted as 20 Ω rather than with extra digits.

Question 27Current Electricity

In a metre bridge experiment, the balance point is obtained at 40 cm from the left end when a known resistance of 6 Ω is in the left gap. What is the unknown resistance in the right gap?

  1. 2.4 Ω
  2. 15 Ω
  3. 9 Ω Correct answer.
  4. 4 Ω

Explanation

At balance the ratio of the gap resistances equals the ratio of the balancing lengths, so R_unknown/6 = 60/40. This gives R_unknown = 6 × 1.5 = 9 Ω.

Question 28Current Electricity

Which condition must hold for a Wheatstone bridge to be balanced, so that no current flows through the galvanometer?

  1. The battery emf must equal the galvanometer resistance
  2. All four arms must have equal resistance
  3. The total resistance of the bridge must be zero
  4. The ratio of the two arms on one side must equal the ratio of the two arms on the other side Correct answer.

Explanation

Balance occurs when the potential at the two galvanometer terminals is equal, which requires P/Q = R/S for the four arms. Equal arms are only a special case of this ratio condition.

Question 29Current Electricity

Two identical cells, each of emf 1.5 V and internal resistance 0.5 Ω, are connected in parallel across a 2 Ω resistor. What current flows through the resistor?

  1. 0.6 A Correct answer.
  2. 0.75 A
  3. 1.5 A
  4. 0.67 A

Explanation

Parallel cells of equal emf keep the emf at 1.5 V but halve the internal resistance to 0.25 Ω. The current is 1.5/(2 + 0.25) = 0.6 A.

Question 30Current Electricity

A rheostat is used in a circuit primarily to

  1. convert alternating current into direct current
  2. vary the resistance and hence control the current in the circuit Correct answer.
  3. store charge for later discharge
  4. measure the emf of a cell without drawing current

Explanation

A rheostat is a variable resistor; sliding its contact changes the resistance in circuit, which changes the current for a given applied voltage. It is a control element, not a measuring or converting device.

Question 31Current Electricity

A potential difference of 12 V is applied across a 3 m long wire of uniform cross-section, and the current density is 4 × 10⁵ A/m². What is the resistivity of the wire material?

  1. 3.0 × 10⁻⁵ Ω m
  2. 1.0 × 10⁻⁴ Ω m
  3. 1.0 × 10⁻⁵ Ω m Correct answer.
  4. 4.0 × 10⁻⁵ Ω m

Explanation

The electric field is V/L = 12/3 = 4 V/m. Since E = ρJ, the resistivity is ρ = E/J = 4/(4 × 10⁵) = 1.0 × 10⁻⁵ Ω m.

Question 32Current Electricity

An electric heater rated 1000 W at 250 V is used on a 200 V supply. Assuming its resistance stays constant, what power does it consume?

  1. 1000 W
  2. 800 W
  3. 1250 W
  4. 640 W Correct answer.

Explanation

The resistance is V²/P = 250²/1000 = 62.5 Ω. On 200 V the power is V²/R = 200²/62.5 = 640 W, which scales as the square of the voltage ratio.

Question 33Magnetic Effects of Current and Magnetism

A straight conductor carrying a steady current is placed perpendicular to a uniform magnetic field. Which quantity determines the magnitude of the magnetic force acting on the conductor?

  1. The product of the conductor's cross-sectional area and the flux density
  2. The product of current, length in the field, and magnetic flux density Correct answer.
  3. The sum of the current and the magnetic flux density
  4. The ratio of magnetic flux density to the current

Explanation

For a straight conductor perpendicular to a uniform field, the force is F = BIL, so the magnitude depends on the product of current, length within the field, and flux density. The area of the conductor and the ratio B/I do not appear in this relation.

Question 34Magnetic Effects of Current and Magnetism

The magnetic field lines produced by a long straight current-carrying wire, in a plane perpendicular to the wire, are best described as:

  1. Parallel straight lines on one side of the wire
  2. Ellipses with the wire at one focus
  3. Radial lines diverging from the wire
  4. Concentric circles centred on the wire Correct answer.

Explanation

By the right-hand grip rule, the field of a long straight wire forms closed circular loops around the wire. In a perpendicular cross-section these appear as concentric circles centred on the wire, with the direction given by the current sense.

Question 35Magnetic Effects of Current and Magnetism

A charged particle enters a uniform magnetic field with its velocity parallel to the field direction. What is the magnetic force on the particle?

  1. Maximum, directed opposite to the field
  2. Half the value obtained when the velocity is perpendicular
  3. Maximum, directed along the velocity
  4. Zero, because the velocity is parallel to the field Correct answer.

Explanation

The magnetic force on a moving charge is F = qvB sinθ, where θ is the angle between velocity and field. When the velocity is parallel to the field, θ = 0 and sinθ = 0, so the force vanishes.

Question 36Magnetic Effects of Current and Magnetism

In a moving-coil galvanometer, the radial magnetic field is used so that:

  1. The coil experiences a constant torque for a given current at every deflection Correct answer.
  2. The number of turns in the coil can be reduced to one
  3. The restoring torque of the spring disappears
  4. The coil's moment of inertia becomes zero

Explanation

A radial field keeps the plane of the coil parallel to the field lines as it turns, so the torque is proportional to current and independent of deflection angle. This makes the deflection directly proportional to the current, giving a linear scale.

Question 37Magnetic Effects of Current and Magnetism

A solenoid of length 0.50 m has 500 turns and carries a current of 2.0 A. Taking μ₀ = 4π × 10⁻⁷ T·m/A, the magnetic flux density near its centre is approximately:

  1. 2.5 × 10⁻³ T Correct answer.
  2. 1.3 × 10⁻³ T
  3. 5.0 × 10⁻³ T
  4. 2.5 × 10⁻⁴ T

Explanation

For a long solenoid, B = μ₀nI with n = N/L = 500/0.50 = 1000 turns per metre. Thus B = (4π × 10⁻⁷)(1000)(2.0) ≈ 2.5 × 10⁻³ T.

Question 38Magnetic Effects of Current and Magnetism

Two long parallel wires carry currents in the same direction. The force between them is:

  1. Attractive only if the currents are unequal
  2. Zero, because the currents are parallel
  3. Attractive, because the fields between the wires partially cancel Correct answer.
  4. Repulsive, because the fields between the wires reinforce

Explanation

Each wire lies in the magnetic field of the other. For parallel currents the fields in the region between the wires oppose each other, reducing the flux density there, and the resulting force pulls the wires together.

Question 39Magnetic Effects of Current and Magnetism

A charged particle moves in a uniform magnetic field with velocity perpendicular to the field. Which statement about its subsequent motion is correct?

  1. It spirals outward with increasing radius
  2. It moves in a circle at constant speed, with the field doing no work Correct answer.
  3. It moves in a circle while its speed steadily increases
  4. It moves in a straight line with constant acceleration

Explanation

The magnetic force is always perpendicular to velocity, so it changes direction but not speed, providing centripetal force for uniform circular motion. Since force is perpendicular to displacement, the magnetic field does no work on the particle.

Question 40Magnetic Effects of Current and Magnetism

A bar magnet is moved rapidly towards a coil connected to a galvanometer. Compared with moving it slowly through the same displacement, the rapid motion produces:

  1. A smaller induced emf, because the magnet spends less time near the coil
  2. The same induced emf, because the flux change is identical
  3. No induced emf, because the magnet's field is static
  4. A larger induced emf, because the rate of change of flux linkage is greater Correct answer.

Explanation

Faraday's law states that the induced emf equals the negative rate of change of magnetic flux linkage. The total flux change is the same, but a shorter time interval makes the rate larger, so the emf is larger for the rapid motion.

Question 41Magnetic Effects of Current and Magnetism

Lenz's law gives the direction of an induced current. It states that the induced current:

  1. Flows in the same sense as the change in flux
  2. Opposes the change in magnetic flux that produced it Correct answer.
  3. Is always clockwise when viewed from above the coil
  4. Depends only on the resistance of the circuit

Explanation

Lenz's law is a statement of energy conservation: the induced current sets up a magnetic field that opposes the flux change causing it. This opposition is why work must be done to drive the change.

Question 42Magnetic Effects of Current and Magnetism

An electron enters a uniform magnetic field of 0.20 T with a speed of 3.0 × 10⁶ m/s perpendicular to the field. Taking e = 1.6 × 10⁻¹⁹ C, the radius of its circular path is approximately:

  1. 4.7 × 10⁻³ m
  2. 8.5 × 10⁻⁵ m Correct answer.
  3. 2.4 × 10⁻⁴ m
  4. 1.1 × 10⁻⁵ m

Explanation

The magnetic force provides centripetal force: evB = mv²/r, so r = mv/(eB). Using the electron mass 9.1 × 10⁻³¹ kg gives r ≈ (9.1 × 10⁻³¹ × 3.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.20) ≈ 8.5 × 10⁻⁵ m.

Question 43Magnetic Effects of Current and Magnetism

A current-carrying coil is placed in a uniform magnetic field with its plane parallel to the field. The torque on the coil is:

  1. Independent of the current in the coil
  2. Half the maximum value
  3. Zero, because the plane of the coil is parallel to the field
  4. Maximum, because the normal to the coil is perpendicular to the field Correct answer.

Explanation

Torque on a coil is τ = NIAB sinθ, where θ is the angle between the coil's normal and the field. With the plane parallel to the field, the normal is perpendicular to it, so θ = 90° and sinθ = 1, giving maximum torque.

Question 44Magnetic Effects of Current and Magnetism

The magnetic field at a point on the axis of a circular current loop is directed along the axis. If the current in the loop is reversed, the field at that point:

  1. Rotates through 90° in the plane of the loop
  2. Reverses direction while keeping the same magnitude Correct answer.
  3. Keeps the same direction but doubles in magnitude
  4. Becomes zero

Explanation

The field direction on the axis is given by the right-hand rule and is proportional to the current. Reversing the current reverses the field vector, while the magnitude, depending on |I|, is unchanged.

Question 45Magnetic Effects of Current and Magnetism

A transformer has 2000 turns in the primary and 100 turns in the secondary. If the primary is connected to a 240 V alternating supply, the secondary emf (assuming an ideal transformer) is:

  1. 12 V Correct answer.
  2. 1200 V
  3. 48 V
  4. 24 V

Explanation

For an ideal transformer, Vs/Vp = Ns/Np. Hence Vs = 240 × (100/2000) = 12 V. The step-down ratio equals the turns ratio.

Question 46Magnetic Effects of Current and Magnetism

A coil of 50 turns and area 0.020 m² lies with its plane perpendicular to a uniform magnetic field that decreases steadily from 0.40 T to zero in 0.10 s. The average induced emf is:

  1. 0.40 V
  2. 0.0040 V
  3. 0.080 V
  4. 4.0 V Correct answer.

Explanation

By Faraday's law, the average emf equals N times the rate of change of flux: emf = N·A·(ΔB/Δt). Here N = 50, A = 0.020 m², and ΔB/Δt = 0.40/0.10 = 4.0 T/s, giving emf = 50 × 0.020 × 4.0 = 4.0 V.

Question 47Magnetic Effects of Current and Magnetism

A charged particle enters a region containing crossed uniform electric and magnetic fields and passes through undeflected. If the electric field is E and the magnetic flux density is B, the speed of the particle must satisfy:

  1. v = B/E, the reciprocal of the field ratio
  2. v = E/(qB), depending on the charge
  3. v = EB, proportional to the product of the fields
  4. v = E/B, independent of the charge and mass Correct answer.

Explanation

For no deflection, the electric force qE and magnetic force qvB must be equal and opposite. Equating them gives qE = qvB, so v = E/B; the charge cancels, and the result is independent of q and m.

Question 48Magnetic Effects of Current and Magnetism

A soft-iron core is inserted into a solenoid carrying a steady current, and the current is kept unchanged. Which statement best explains the resulting increase in magnetic flux density inside the solenoid?

  1. The iron becomes magnetised and its aligned domains add their own field to the solenoid's field Correct answer.
  2. The iron reduces the number of turns per metre, concentrating the field
  3. The iron removes the field of the solenoid and replaces it with a purely electric field
  4. The iron raises the current by reducing the resistance of the coil

Explanation

Soft iron has high relative permeability, so its domains align with the applied field and produce an additional internal field. The total flux density B = μ₀μᵣnI is therefore much larger than for an air core at the same current.

Question 49Electromagnetic Induction and Alternating Currents

A bar magnet is pushed north pole first into a coil connected to a galvanometer. According to Lenz's law, the induced current in the coil will produce a magnetic field at the face nearest the magnet that

  1. alternates direction as the magnet moves
  2. points away from the magnet, aiding its approach
  3. points toward the magnet, opposing its approach Correct answer.
  4. is zero because the flux is increasing

Explanation

Lenz's law states that the induced current opposes the change producing it. As the north pole approaches, flux through the coil increases, so the induced field at the near face must be a north pole, i.e. directed toward the magnet, repelling it. This is a direct consequence of energy conservation.

Question 50Electromagnetic Induction and Alternating Currents

A coil of 200 turns and area 0.05 m² lies in a uniform magnetic field that is perpendicular to the plane of the coil. The field is reduced uniformly from 0.4 T to 0.1 T in 0.2 s. The magnitude of the induced EMF in the coil is

  1. 30 V
  2. 15 V Correct answer.
  3. 60 V
  4. 5 V

Explanation

Faraday's law gives EMF = N·ΔΦ/Δt, with Φ = BA cosθ and θ = 0. The change in flux is 0.05 × (0.4 − 0.1) = 0.015 Wb. Multiplying by 200 turns and dividing by 0.2 s gives 15 V.