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Assam Higher Secondary Physics study materials
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Assam Higher Secondary Physics practice questions

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Question 1Electrostatics

Two point charges of +3 μC and −3 μC are separated by 0.20 m in air. What is the magnitude of the electrostatic force between them? (k = 9 × 10⁹ N m² C⁻²)

  1. 4.5 N
  2. 2.0 N Correct answer.
  3. 9.0 N
  4. 0.45 N

Explanation

Coulomb's law gives F = k|q₁q₂|/r². Substituting, F = (9 × 10⁹ × 3 × 10⁻⁶ × 3 × 10⁻⁶)/(0.20)² = (8.1 × 10⁻²)/0.04 = 2.0 N. The signs only fix the direction (attractive), not the magnitude.

Question 2Electrostatics

The electric field at a point 0.30 m from a point charge of 2.0 × 10⁻⁸ C in air is

  1. 6.0 × 10² N C⁻¹
  2. 2.0 × 10³ N C⁻¹ Correct answer.
  3. 1.8 × 10³ N C⁻¹
  4. 6.0 × 10³ N C⁻¹

Explanation

For a point charge the field magnitude is E = kq/r². Here E = (9 × 10⁹ × 2.0 × 10⁻⁸)/(0.30)² = 180/0.09 = 2.0 × 10³ N C⁻¹. The field points radially away from the charge because it is positive.

Question 3Electrostatics

The dimensional formula of electric flux is

  1. [M L³ T⁻³ A⁻¹] Correct answer.
  2. [M L² T⁻³ A⁻¹]
  3. [M L² T⁻² A⁻¹]
  4. [M L³ T⁻² A⁻¹]

Explanation

Electric flux is E·A, and E has dimensions [M L T⁻³ A⁻¹] while area has [L²]. Multiplying gives [M L³ T⁻³ A⁻¹], which is also the dimensional form of charge divided by permittivity.

Question 4Electrostatics

A hollow conducting sphere carries a net charge. At a point inside the cavity, well away from the inner surface, the electric field due to the charge is

  1. directed radially outward
  2. equal to the field at the surface
  3. directed radially inward
  4. zero Correct answer.

Explanation

In electrostatic equilibrium the free charges reside on the outer surface, and the field inside the conducting material and the empty cavity is zero. This follows from Gauss's law applied to a Gaussian surface inside the conductor enclosing no net charge.

Question 5Electrostatics

The work done in moving a charge of 5 μC between two points whose potentials differ by 40 V is

  1. 1.2 × 10⁻³ J
  2. 2.0 × 10⁻³ J
  3. 2.0 × 10⁻⁴ J Correct answer.
  4. 8.0 × 10⁻⁴ J

Explanation

The work done by the external agent equals q times the potential difference, W = qΔV. Thus W = 5 × 10⁻⁶ × 40 = 2.0 × 10⁻⁴ J, independent of the path taken.

Question 6Electrostatics

Three capacitors of 2 μF, 3 μF and 6 μF are joined in series. The equivalent capacitance of the combination is

  1. 1.5 μF
  2. 0.5 μF
  3. 11 μF
  4. 1.0 μF Correct answer.

Explanation

For capacitors in series, 1/C = 1/C₁ + 1/C₂ + 1/C₃. Here 1/C = 1/2 + 1/3 + 1/6 = 1, so C = 1.0 μF, which is smaller than the smallest individual capacitor.

Question 7Electrostatics

A parallel-plate capacitor with air between its plates has capacitance C. A dielectric of relative permittivity 4 completely fills the gap while the capacitor stays connected to the same battery. The stored energy becomes

  1. C V²/8
  2. C V²/2
  3. 4 C V²
  4. 2 C V² Correct answer.

Explanation

With the battery connected, the potential difference V stays fixed while the capacitance rises to 4C. Energy is ½CV², so it becomes ½(4C)V² = 2CV², four times the original ½CV².

Question 8Electrostatics

An electric dipole of moment p is placed in a uniform field E with its axis at 30° to the field. The magnitude of the torque acting on it is

  1. pE√3/2
  2. zero
  3. pE
  4. pE/2 Correct answer.

Explanation

The torque on a dipole is τ = pE sin θ. With θ = 30°, sin 30° = 0.5, so the torque is pE/2. The torque is maximum at 90° and zero when the dipole is aligned with the field.

Question 9Electrostatics

The electric potential due to a short dipole at a point on its axial line at distance r is V. At the same distance on the equatorial line, the potential is

  1. zero Correct answer.
  2. V
  3. −V/2
  4. V/2

Explanation

On the equatorial line every point is equidistant from the two equal and opposite charges of the dipole, so their potentials cancel exactly. The potential there is zero, whereas on the axial line it is kp/r².

Question 10Electrostatics

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/ε₀
  2. Q/12ε₀
  3. Q/6ε₀ Correct answer.
  4. Q/8ε₀

Explanation

By Gauss's law the total flux through the closed surface is Q/ε₀. Because the charge is at the centre, symmetry makes the flux through all six faces equal, so each face carries Q/6ε₀.

Question 11Electrostatics

Two identical small conducting spheres carry charges +6 μC and −2 μC. They are touched together and then separated to the same distance. The ratio of the new force to the original force is

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

Explanation

On contact the total charge +4 μC is shared equally, giving +2 μC on each sphere. The original force product is 12 μC² and the new product is 4 μC², so the ratio is 4:12 = 1:3.

Question 12Electrostatics

The electric field just outside the surface of a charged conductor is 4.0 × 10⁴ N C⁻¹. The surface charge density on the conductor is closest to (ε₀ = 8.85 × 10⁻¹² F m⁻¹)

  1. 2.2 × 10⁻⁷ C m⁻²
  2. 4.4 × 10⁻⁷ C m⁻²
  3. 3.5 × 10⁻⁷ C m⁻² Correct answer.
  4. 1.1 × 10⁻⁷ C m⁻²

Explanation

At the surface of a conductor the field is E = σ/ε₀, so σ = ε₀E. Substituting gives σ = 8.85 × 10⁻¹² × 4.0 × 10⁴ ≈ 3.5 × 10⁻⁷ C m⁻².

Question 13Electrostatics

A capacitor of capacitance C is charged to potential V and then disconnected from the battery. The plates are now pulled apart so the separation doubles. Which statement about the stored energy is correct?

  1. It doubles because the capacitance halves at constant charge. Correct answer.
  2. It becomes four times larger because the field doubles.
  3. It stays the same because charge is conserved.
  4. It halves because the capacitance doubles at constant charge.

Explanation

After disconnection the charge Q is fixed, and energy is U = Q²/2C. Doubling the separation halves the capacitance, so the energy doubles; the external agent does work against the attraction of the plates.

Question 14Electrostatics

Two point charges, +q and +4q, are fixed a distance L apart. At what distance from the +q charge on the line joining them is the net electric field zero?

  1. 3L/5
  2. L/4
  3. 2L/3
  4. L/3 Correct answer.

Explanation

The null point lies between the charges, where the fields oppose. Setting kq/x² = k(4q)/(L − x)² gives (L − x)/x = 2, hence x = L/3, closer to the smaller charge.

Question 15Electrostatics

A spherical shell of radius R carries a uniform surface charge density σ. A point P lies at distance r < R from the centre. Which statement about the potential at P is correct?

  1. It is zero because the field inside is zero.
  2. It equals σr/ε₀ and falls linearly to zero at the centre.
  3. It equals σR/ε₀ and varies with r.
  4. It equals σR/ε₀ and is the same as at the surface. Correct answer.

Explanation

Inside a uniformly charged shell the field is zero, so the potential does not change with position and equals its surface value σR/ε₀. Zero field means constant potential, not zero potential.

Question 16Electrostatics

A 2 μF capacitor is charged to 100 V and then connected in parallel with an uncharged 3 μF capacitor. The common potential attained and the energy lost in the process are

  1. 40 V and 6 mJ Correct answer.
  2. 40 V and 12 mJ
  3. 60 V and 12 mJ
  4. 60 V and 6 mJ

Explanation

Charge is conserved: 2 μF × 100 V = 200 μC is shared over 5 μF, giving V = 40 V. Initial energy is 10 mJ and final energy is ½ × 5 μF × (40 V)² = 4 mJ, so 6 mJ is lost as heat and radiation.

Question 17Current Electricity

A uniform wire of resistance R is stretched so that its new length is twice the original length, its volume remaining unchanged. What is the resistance of the stretched wire?

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

Explanation

When a wire is stretched at constant volume, its length doubles while its cross-sectional area halves. Resistance equals resistivity times length divided by area, so doubling the length and halving the area multiplies the resistance by four. Hence the new resistance is 4R.

Question 18Current Electricity

The drift velocity of free electrons in a metallic conductor carrying a steady current is directly proportional to which quantity?

  1. The number density of free electrons in the metal
  2. The cross-sectional area of the conductor
  3. The square of the conductor's length
  4. The applied potential difference across the conductor Correct answer.

Explanation

Drift velocity is given by v = I/(neA), and since current is proportional to the potential difference for an ohmic conductor, drift velocity rises with the applied voltage. It is inversely related to both the free-electron density and the cross-sectional area. Length affects resistance and hence current, but the direct proportionality asked for is with potential difference.

Question 19Current Electricity

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

  1. 0.5 ohm
  2. 1.5 ohm
  3. 1 ohm Correct answer.
  4. 11 ohm

Explanation

For a parallel combination the reciprocals add: 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. Therefore the equivalent resistance is 1 ohm, which is smaller than the smallest individual resistor as expected for a parallel network.

Question 20Current Electricity

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

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

Explanation

The junction rule states that the total current entering a node equals the total current leaving it, which expresses the fact that charge cannot accumulate at a point in a steady circuit. The loop rule, by contrast, follows from conservation of energy.

Question 21Current Electricity

A battery of emf 12 V and internal resistance 1 ohm is connected to an external resistance of 5 ohm. What is the terminal potential difference of the battery?

  1. 10 V Correct answer.
  2. 12 V
  3. 7 V
  4. 2 V

Explanation

The current is emf divided by total resistance, giving 12/6 = 2 A. The terminal voltage equals emf minus the internal drop, so 12 - (2 x 1) = 10 V. This also equals the current times the external resistance, 2 x 5 = 10 V.

Question 22Current Electricity

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

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

Explanation

Two diametrically opposite points divide the circle into two semicircular halves, each of resistance 5 ohm, connected in parallel. The parallel combination of two equal 5 ohm resistors is 2.5 ohm.

Question 23Current Electricity

The temperature coefficient of resistance of a metal is positive. As the temperature of a metallic conductor rises, what happens to its resistance and to its resistivity?

  1. Both resistance and resistivity increase Correct answer.
  2. Resistance increases while resistivity stays constant
  3. Both resistance and resistivity decrease
  4. Resistance stays constant while resistivity increases

Explanation

In a metal, increased temperature raises lattice vibrations, which increases the frequency of collisions of free electrons and shortens their mean free time. This increases resistivity, and since resistance equals resistivity times length divided by area, resistance also rises.

Question 24Current Electricity

A galvanometer of resistance 50 ohm gives full-scale deflection at 1 mA. What shunt resistance is needed to convert it into an ammeter reading up to 10 mA?

  1. 500 ohm
  2. 5.56 ohm Correct answer.
  3. 0.5 ohm
  4. 50 ohm

Explanation

The shunt carries the excess current, 10 mA - 1 mA = 9 mA, while the galvanometer carries 1 mA at full scale. Since the shunt and galvanometer are in parallel they have equal voltage, so S x 9 = 50 x 1, giving S = 50/9, about 5.56 ohm.

Question 25Current Electricity

A cell of emf 2 V and internal resistance 0.5 ohm is connected to a variable external resistance. At what external resistance is the power delivered to the external circuit maximum?

  1. 1 ohm
  2. 2 ohm
  3. 0.5 ohm Correct answer.
  4. 0 ohm

Explanation

The power delivered to the external resistor is maximum when the external resistance equals the internal resistance of the source. This is the condition for maximum power transfer, so the external resistance should be 0.5 ohm.

Question 26Current Electricity

Which relation correctly expresses the current density J in a conductor in terms of the electric field E?

  1. J equals sigma squared times E
  2. J equals sigma times E, where sigma is conductivity Correct answer.
  3. J equals E divided by sigma
  4. J equals sigma times E squared

Explanation

In vector form Ohm's law at a point is J = sigma E, where sigma is the conductivity of the material. This is the microscopic form of Ohm's law, and its reciprocal relation gives resistivity as E/J.

Question 27Current Electricity

A carbon resistor and a metallic wire are connected in series and the combination is heated. If the carbon resistor has a small negative temperature coefficient and the metal has a positive one, what is the effect on the total resistance if the metal's change dominates?

  1. The total resistance becomes zero
  2. The total resistance remains exactly unchanged
  3. The total resistance increases Correct answer.
  4. The total resistance decreases

Explanation

The metal's resistance rises with temperature while the carbon resistor's resistance falls slightly. Since the metal's positive change is stated to dominate, the net increase from the metal outweighs the small decrease from carbon, so the total resistance increases.

Question 28Current Electricity

In a potentiometer experiment, a cell gives a balance point at 60 cm of the wire. When a resistor of 5 ohm is connected across the cell, the balance point shifts to 50 cm. What is the internal resistance of the cell?

  1. 5 ohm
  2. 0.5 ohm
  3. 6 ohm
  4. 1 ohm Correct answer.

Explanation

Internal resistance is r = R (L1 - L2)/L2, where L1 is the open-circuit balance length and L2 the balance length with the resistor. Substituting gives r = 5 x (60 - 50)/50 = 5 x 0.2 = 1 ohm.

Question 29Current Electricity

Two cells of emf 6 V and 4 V with internal resistances 1 ohm and 2 ohm respectively are connected in parallel with their like terminals together. What is the terminal voltage of the combination when no external load is connected?

  1. 4 V
  2. 5.33 V Correct answer.
  3. 10 V
  4. 5 V

Explanation

For cells in parallel with no external load, the terminal voltage is the net emf divided by the total internal conductance, giving (6/1 + 4/2)/(1/1 + 1/2) = (6 + 2)/(1.5) = 8/1.5, about 5.33 V. This is the potential across the parallel combination.

Question 30Current Electricity

A heater rated 1000 W at 220 V is operated at 110 V. Assuming its resistance stays constant, what power does it consume?

  1. 2000 W
  2. 1000 W
  3. 500 W
  4. 250 W Correct answer.

Explanation

The resistance is R = V squared / P = 220 squared / 1000 = 48.4 ohm. At 110 V the power is V squared / R = 110 squared / 48.4 = 250 W. Halving the voltage quarters the power for a fixed resistance.

Question 31Current Electricity

A Wheatstone bridge has arms P and Q in the upper branches and R and S in the lower branches, with a galvanometer between the midpoints. The bridge is balanced. If P and Q are interchanged while R and S remain fixed, what is the new state of the bridge?

  1. The bridge becomes unbalanced and the balance condition changes to Q/P equals R/S Correct answer.
  2. The bridge becomes balanced only if the battery is reversed
  3. The bridge can never be rebalanced by any change of arms
  4. The bridge remains balanced because the ratio condition is symmetric

Explanation

Balance requires the ratio of the upper arms to equal the ratio of the lower arms, P/Q = R/S. Interchanging P and Q inverts the left-hand ratio, so the condition becomes Q/P = R/S, which generally fails. The bridge therefore goes out of balance unless the arms happen to be equal.

Question 32Current Electricity

A set of n identical cells, each of emf E and internal resistance r, is connected in series with an external resistance R. What value of R gives maximum power in the external resistor?

  1. nr Correct answer.
  2. r
  3. r/n
  4. n squared times r

Explanation

For n identical cells in series, the total emf is nE and the total internal resistance is nr. Maximum power transfer to the external resistor occurs when R equals the total internal resistance, so R = nr.

Question 33Magnetic Effects of Current and Magnetism

A straight wire carries a steady current I. According to the Biot–Savart law, the magnetic field at a point on the axis of the wire, far from its ends, is best described as:

  1. Directed radially outward from the wire
  2. Directed along the wire, parallel to the current
  3. Infinite, because the distance from every element is small
  4. Zero, because the position vector is parallel to the current element Correct answer.

Explanation

The Biot–Savart contribution is proportional to dl × r̂, and on the axis dl is parallel to r, so the cross product vanishes for every element. Hence the net field on the axis of an ideal straight current-carrying wire is zero. Off-axis points, by contrast, give the familiar circular field pattern.

Question 34Magnetic Effects of Current and Magnetism

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

  1. Attractive, since the magnetic fields reinforce between them Correct answer.
  2. Repulsive, since like currents repel
  3. Attractive only if the currents are unequal in magnitude
  4. Zero, since the currents are parallel

Explanation

Each wire produces a magnetic field that, at the location of the other, points so that the force qv × B draws the wires together. Parallel currents therefore attract, and antiparallel currents repel. The magnitude per unit length is μ₀I₁I₂/(2πd) regardless of whether the currents are equal.

Question 35Magnetic Effects of Current and Magnetism

A charged particle enters a uniform magnetic field with its velocity perpendicular to the field. Which quantity remains constant during the motion inside the field?

  1. The speed of the particle Correct answer.
  2. The kinetic energy and the velocity vector both change direction together
  3. The velocity vector of the particle
  4. The momentum vector of the particle

Explanation

The magnetic force qv × B is always perpendicular to the velocity, so it does no work on the particle. The kinetic energy and hence the speed stay constant while only the direction of motion changes. The velocity and momentum vectors rotate continuously, so they are not constant.

Question 36Magnetic 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. μ₀I/(2πr)
  2. μ₀NI/(2πr)
  3. μ₀NI/(2r) Correct answer.
  4. μ₀NI/(4πr)

Explanation

Each turn contributes μ₀I/(2r) at the centre, and N turns in series add their fields in the same direction. The result is B = μ₀NI/(2r). The 2πr forms belong to the field of a long straight wire, not a coil centre.

Question 37Magnetic Effects of Current and Magnetism

A bar magnet is cut into two pieces perpendicular to its axis. Each piece behaves as:

  1. A magnet with only a south pole
  2. A non-magnetic fragment
  3. A complete magnet with both a north and a south pole Correct answer.
  4. A magnet with only a north pole

Explanation

Magnetic poles always exist in pairs; a magnetic monopole has never been observed. Cutting a magnet simply creates two smaller dipoles, each with its own north and south pole. This is a direct consequence of the absence of free magnetic charge.

Question 38Magnetic Effects of Current and Magnetism

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

  1. 0.025 Ω
  2. 0.050 Ω Correct answer.
  3. 50 Ω
  4. 0.50 Ω

Explanation

A shunt S carries the excess current: I_g G = (I − I_g) S. Here S = (2×10⁻³ × 50)/(2 − 0.002) ≈ 0.1/1.998 ≈ 0.050 Ω. A low shunt diverts most of the current so the meter reads the full line current.

Question 39Magnetic Effects of Current and Magnetism

A galvanometer of resistance G and full-scale current I_g is converted into a voltmeter of range V by connecting a series resistance R. Which relation gives R?

  1. R = V/I_g + G
  2. R = V/(I_g·G)
  3. R = V·G/I_g
  4. R = V/I_g − G Correct answer.

Explanation

For full-scale deflection the current through the series combination must be I_g when the terminal voltage is V. Thus V = I_g(G + R), giving R = V/I_g − G. The series resistance limits current and drops the excess voltage.

Question 40Magnetic Effects of Current and Magnetism

A circular loop of radius 0.10 m carries a current of 2.0 A. The magnitude of the magnetic dipole moment of the loop is:

  1. 0.063 A·m² Correct answer.
  2. 0.020 A·m²
  3. 0.20 A·m²
  4. 0.63 A·m²

Explanation

The magnetic moment is m = I·A, with A = πr² = π(0.10)² ≈ 0.0314 m². Multiplying by I = 2.0 A gives m ≈ 0.063 A·m². The direction is along the axis given by the right-hand rule.

Question 41Magnetic Effects of Current and Magnetism

An electron moves with speed v in a circle of radius r perpendicular to a uniform magnetic field B. If the speed is doubled while B is unchanged, the radius of the circular path becomes:

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

Explanation

The magnetic force supplies the centripetal force: evB = mv²/r, so r = mv/(eB). For fixed B and fixed particle, r is directly proportional to v. Doubling the speed therefore doubles the radius.

Question 42Magnetic Effects of Current and Magnetism

A current-carrying conductor of length L is placed in a uniform magnetic field B at an angle θ to the field. The magnitude of the magnetic force on it is:

  1. BIL, independent of θ
  2. BIL sin θ Correct answer.
  3. BIL cos θ
  4. BIL tan θ

Explanation

The force on a straight conductor is F = I L × B, whose magnitude is BIL sin θ, where θ is the angle between the current direction and the field. The force is maximum at 90° and zero when the conductor lies along the field.

Question 43Magnetic Effects of Current and Magnetism

Two infinitely long straight wires carry currents I and 2I in opposite directions and are separated by distance d. The magnetic field is zero at a point:

  1. On the side of the wire carrying 2I, at distance d from it
  2. Nowhere, since the fields always add
  3. On the side of the wire carrying I, at distance d from it Correct answer.
  4. Exactly midway between the wires

Explanation

Between the wires the fields of opposite currents point the same way, so they add and cannot cancel. Outside, on the side of the smaller current, the fields oppose; setting μ₀I/(2πx) = μ₀(2I)/(2π(x+d)) gives x = d from the wire carrying I.

Question 44Magnetic Effects of Current and Magnetism

A bar magnet is moved quickly towards a closed coil and then held stationary inside it. During the interval when the magnet is stationary, the induced current in the coil is:

  1. Large and steady
  2. Oscillating with the magnet's motion history
  3. Equal to the current induced during the approach
  4. Zero Correct answer.

Explanation

Faraday's law states that the induced emf equals the negative rate of change of magnetic flux. While the magnet is stationary the flux through the coil is constant, so the rate of change is zero and no current flows. Current appears only while the flux is changing.

Question 45Magnetic Effects of Current and Magnetism

A charged particle enters a region containing both a uniform electric field E and a uniform magnetic field B, with E perpendicular to B. It passes undeviated when its speed satisfies:

  1. v = 1/(EB)
  2. v = E/B Correct answer.
  3. v = EB
  4. v = B/E

Explanation

For no deflection the electric force qE and the magnetic force qvB must be equal and opposite. Setting qE = qvB gives v = E/B, the velocity selected by a velocity selector. Particles with other speeds are deflected.

Question 46Magnetic Effects of Current and Magnetism

A sensitive moving-coil galvanometer is connected in series with a high resistance and used as a voltmeter. Which change would increase its voltage range without exceeding full-scale deflection current?

  1. Replace the magnet with a weaker one
  2. Decrease the series resistance
  3. Increase the series resistance Correct answer.
  4. Decrease the number of turns in the coil

Explanation

Increasing the series multiplier resistance reduces the current for a given applied voltage, so a larger voltage is needed to reach the full-scale current I_g. The new range is V = I_g(G + R). Reducing the resistance or weakening the coil's response would lower the range instead.

Question 47Magnetic Effects of Current and Magnetism

A long solenoid has n turns per unit length and carries current I. A second, shorter solenoid is wound closely over the middle of the first, with the same number of turns per unit length, and carries the same current in the opposite sense. The magnetic field inside the overlap region is:

  1. Zero Correct answer.
  2. μ₀nI/2, directed opposite to the first solenoid's field
  3. 2μ₀nI, directed along the axis
  4. μ₀nI, directed along the axis

Explanation

Inside a long solenoid the field is μ₀nI along the axis, and the short over-wound solenoid produces an equal field in the opposite direction in the overlap. The two contributions cancel exactly, leaving zero field there. Outside the short solenoid, the long solenoid's field is essentially unchanged.

Question 48Magnetic Effects of Current and Magnetism

A rectangular coil of N turns, area A, carrying current I is held in a uniform magnetic field B with its plane parallel to the field. The magnitude of the torque on the coil is:

  1. Zero, because the normal is perpendicular to B
  2. 2NIAB
  3. NIAB, because the normal is perpendicular to B Correct answer.
  4. NIAB/2

Explanation

The torque is τ = m × B with m = NIA along the coil normal. When the coil plane is parallel to B, the normal is perpendicular to B, so sin θ = 1 and the torque magnitude is NIAB, its maximum value. The torque vanishes when the plane is perpendicular to the field.

Question 49Electromagnetic Induction and Alternating Currents

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

  1. 0.04 V
  2. 0.8 V
  3. 4.0 V
  4. 0.4 V Correct answer.

Explanation

For a rod moving perpendicular to the field, the motional emf is e = Bvl. Substituting B = 0.2 T, v = 4 m/s and l = 0.5 m gives e = 0.2 × 4 × 0.5 = 0.4 V.

Question 50Electromagnetic 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 charge
  4. Conservation of linear momentum

Explanation

Lenz's law states that the induced current opposes the change producing it. If it aided the change, the induced current would grow without external work, so the law follows from conservation of energy.