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PSEB SS Physics practice questions

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Question 1Electrostatics

Two point charges separated by a distance r experience an electrostatic force F. If the separation is increased to 3r while the charges are left unchanged, the magnitude of the force becomes:

  1. F/9 Correct answer.
  2. 9F
  3. F/3
  4. 3F

Explanation

Coulomb's law gives F = kq₁q₂/r², so the force between two point charges is inversely proportional to the square of their separation. Replacing r by 3r multiplies the force by 1/9, so the new force is F/9.

Question 2Electrostatics

In SI units, electric flux through a surface is measured in:

  1. N C⁻¹
  2. C m⁻²
  3. N m² C⁻¹ Correct answer.
  4. N m C⁻¹

Explanation

Electric flux is defined as the product of electric field strength and area, so its unit is the unit of field (N C⁻¹) multiplied by the unit of area (m²). This gives N m² C⁻¹, which is equivalently written as V m.

Question 3Electrostatics

A solid metallic conductor carrying a net charge is in electrostatic equilibrium. At any point inside the material of the conductor, the net electric field is:

  1. zero Correct answer.
  2. directed radially outward from the centre
  3. uniform and non-zero
  4. equal to the field just outside its surface

Explanation

In equilibrium the free electrons of the conductor redistribute until the field they produce exactly cancels any applied field inside the material. The excess charge then resides on the outer surface, where the field is normal to the surface and has magnitude σ/ε₀.

Question 4Electrostatics

An electric dipole of moment 4.0 × 10⁻⁹ C m is placed in a uniform electric field of 5.0 × 10⁴ N C⁻¹ with its axis making an angle of 30° with the field. The magnitude of the torque on the dipole is:

  1. 2.0 × 10⁻⁴ N m
  2. 4.0 × 10⁻⁴ N m
  3. 1.0 × 10⁻⁴ N m Correct answer.
  4. 1.73 × 10⁻⁴ N m

Explanation

The torque on a dipole in a uniform field is τ = pE sinθ. Here pE = (4.0 × 10⁻⁹)(5.0 × 10⁴) = 2.0 × 10⁻⁴ N m and sin 30° = 0.5, so τ = 1.0 × 10⁻⁴ N m.

Question 5Electrostatics

A point 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/(8ε₀)
  2. q/ε₀
  3. q/(6ε₀) Correct answer.
  4. q/(2ε₀)

Explanation

By Gauss's law the total flux through the closed surface of the cube is q/ε₀, because the charge enclosed is q. The charge is at the centre, so all six faces are equivalent by symmetry and each receives one sixth of the total flux.

Question 6Electrostatics

An isolated charged parallel-plate capacitor carries charge Q and has potential difference V across its plates. A dielectric slab of dielectric constant K = 4 is then inserted so that it completely fills the gap. The new potential difference across the plates is:

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

Explanation

Because the capacitor is isolated, its charge Q stays constant while the dielectric raises the capacitance from C to KC = 4C. Since V = Q/C, increasing the capacitance four times reduces the potential difference to one quarter of its original value.

Question 7Electrostatics

The energy stored per unit volume in a region of free space where the electric field has magnitude 3.0 × 10⁴ V m⁻¹ is (ε₀ = 8.85 × 10⁻¹² F m⁻¹):

  1. 7.97 × 10⁻³ J m⁻³
  2. 3.98 × 10⁻⁴ J m⁻³
  3. 3.98 × 10⁻³ J m⁻³ Correct answer.
  4. 1.99 × 10⁻³ J m⁻³

Explanation

The energy density of an electric field in vacuum is u = ½ε₀E². Substituting the values gives u = 0.5 × 8.85 × 10⁻¹² × (3.0 × 10⁴)² = 3.98 × 10⁻³ J m⁻³.

Question 8Electrostatics

A 3 µF capacitor and a 6 µF capacitor are connected in series across a 12 V battery. The charge stored on the 3 µF capacitor is:

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

Explanation

For a series combination the equivalent capacitance is C₁C₂/(C₁ + C₂) = (3 × 6)/9 = 2 µF. The charge on each capacitor in series is the same and equals CV = 2 µF × 12 V = 24 µC.

Question 9Electrostatics

Two isolated conducting spheres of radii R and 2R, initially carrying charge Q each, are joined by a long thin conducting wire. After equilibrium is reached, which statement about the spheres is correct?

  1. The smaller sphere has twice the surface charge density of the larger sphere. Correct answer.
  2. The electric potential of the larger sphere is twice that of the smaller sphere.
  3. The larger sphere has twice the surface charge density of the smaller sphere.
  4. The two spheres carry equal charge.

Explanation

The wire makes the two spheres a single conductor, so their potentials become equal: kQ₁/R = kQ₂/(2R), which gives Q₂ = 2Q₁ and shares the total charge 2Q as 2Q/3 on the small sphere and 4Q/3 on the large one. Since σ = Q/(4πR²), the small sphere has σ = Q/(6πR²) while the larger has σ = Q/(12πR²), so the smaller sphere's surface charge density is twice as large.

Question 10Electrostatics

An electron moving horizontally with speed 1.0 × 10⁷ m s⁻¹ enters a uniform electric field of 1.0 × 10³ N C⁻¹ directed vertically, between two parallel plates 5.0 cm long. The vertical deflection of the electron as it leaves the plates is closest to (e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg):

  1. 0.22 mm
  2. 2.2 mm Correct answer.
  3. 1.1 mm
  4. 4.4 mm

Explanation

The vertical acceleration is a = eE/m = (1.6 × 10⁻¹⁹ × 1.0 × 10³)/(9.1 × 10⁻³¹) ≈ 1.8 × 10¹⁴ m s⁻², while the horizontal motion is uniform, giving t = L/v = 0.050/1.0 × 10⁷ = 5.0 × 10⁻⁹ s. The deflection is then y = ½at² ≈ 0.5 × 1.8 × 10¹⁴ × (5.0 × 10⁻⁹)² ≈ 2.2 × 10⁻³ m, that is about 2.2 mm.

Question 11Electrostatics

A glass rod becomes positively charged when it is rubbed with silk. This happens because the rod

  1. loses electrons to the silk Correct answer.
  2. gains protons from the silk
  3. gains electrons from the silk
  4. shares its protons with the silk

Explanation

When glass and silk are rubbed, electrons are transferred from the glass to the silk because silk holds electrons more strongly. The glass is left with a deficiency of electrons, so it carries a net positive charge. Protons are bound in the nuclei and are not transferred during ordinary rubbing.

Question 12Electrostatics

Which statement about electrostatic field lines is correct?

  1. They meet equipotential surfaces at right angles. Correct answer.
  2. They begin on negative charges and end on positive charges.
  3. They may intersect where the field is zero.
  4. They are always straight lines between two charges.

Explanation

Electric field lines point from positive charge to negative charge, so beginning on negative charges reverses the direction convention. At any point the field has one definite direction, so two field lines cannot cross, and field lines are generally curved. The field is perpendicular to an equipotential surface everywhere, so the lines meet such surfaces at 90 degrees.

Question 13Electrostatics

A parallel-plate capacitor has plate area A and plate separation d. If the separation is doubled while the area stays the same, the capacitance

  1. becomes twice its original value
  2. remains unchanged
  3. becomes four times larger
  4. is reduced to one half Correct answer.

Explanation

For a parallel-plate capacitor in vacuum, C = ε0 A / d. Doubling d while keeping A fixed doubles the denominator, so the capacitance falls to one half. Capacitance depends on the geometry of the plates, so a larger separation stores less charge for the same potential difference.

Question 14Electrostatics

Two small charged spheres separated by a fixed distance repel each other with a force F in vacuum. When the same arrangement is immersed in a liquid of dielectric constant K = 4, the force between the spheres becomes

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

Explanation

In a medium the Coulomb force is reduced by the relative permittivity, or dielectric constant, of that medium, so F_medium = F_vacuum / K. Substituting K = 4 gives F/4. The charges and their separation are unchanged; only the medium between them is different.

Question 15Electrostatics

An isolated point charge produces a potential V at a point that is a distance r from it. The magnitude of the electric field at that same point is

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

Explanation

For an isolated point charge, V = kQ/r and E = kQ/r². Dividing E by V gives E/V = 1/r, so E = V/r at that point. The field is directed along the line joining the charge to the point, outward for positive charge and inward for negative charge.

Question 16Electrostatics

A large uniformly charged plane sheet carries a surface charge density σ. The magnitude of the electric field at a point close to the sheet, at a perpendicular distance x from it, is

  1. σ / (2ε0 x)
  2. σ / (2ε0) Correct answer.
  3. σ / ε0
  4. σ x / (2ε0)

Explanation

Applying Gauss's law to a pillbox that crosses the sheet gives 2EA = σA/ε0, so E = σ/(2ε0). This result contains no x, so the field is uniform on each side of a large sheet. The value σ/ε0 is the field between two oppositely charged sheets, not the field of a single sheet.

Question 17Electrostatics

A parallel-plate capacitor is kept connected to a battery of constant voltage V. If the plate separation is doubled, the charge on the capacitor

  1. becomes twice as large
  2. becomes four times larger
  3. remains the same
  4. becomes half as large Correct answer.

Explanation

Because the battery remains connected, the potential difference V across the plates is fixed. Doubling the separation halves the capacitance, and from Q = CV a halved capacitance at constant V gives half the charge. The electric field between the plates also falls to half, since E = V/d with d doubled.

Question 18Electrostatics

A 4 µF capacitor is charged to a potential difference of 50 V. The energy stored in the capacitor is

  1. 1.0 × 10⁻² J
  2. 5.0 × 10⁻³ J Correct answer.
  3. 2.0 × 10⁻⁴ J
  4. 1.0 × 10⁻⁴ J

Explanation

The energy stored in a capacitor is U = ½CV². Substituting C = 4 × 10⁻⁶ F and V = 50 V gives U = ½ × 4 × 10⁻⁶ × 2500 = 5.0 × 10⁻³ J. Using CV² without the factor ½ would double the answer to 1.0 × 10⁻² J.

Question 19Electrostatics

Charges +q, −q, +q and −q are fixed at the corners A, B, C and D of a square in that order around the perimeter. The net electric field at the centre of the square is

  1. directed along the diagonal AC
  2. directed along the diagonal BD
  3. zero Correct answer.
  4. perpendicular to the plane of the square

Explanation

At the centre, the fields from A (+q) and C (+q) are equal in magnitude and opposite in direction, so they cancel. The fields from B (−q) and D (−q) are also equal and opposite, so they cancel in the same way. All four contributions therefore add to zero, giving no net field at the centre.

Question 20Electrostatics

A 4 µF capacitor charged to 100 V is disconnected from the supply and then connected in parallel with an uncharged 6 µF capacitor. The common potential difference across the combination is

  1. 40 V Correct answer.
  2. 50 V
  3. 100 V
  4. 60 V

Explanation

The initial charge is Q = C1V1 = 4 × 10⁻⁶ × 100 = 4 × 10⁻⁴ C, and this charge is shared between the two capacitors after connection. The total capacitance is 4 µF + 6 µF = 10 µF, so the common potential is V = Q/C_total = 4 × 10⁻⁴ / 10 × 10⁻⁶ = 40 V. Some energy is lost as heat in the connecting wires, but charge is conserved.

Question 21Electrostatics

The statement that the net charge on any isolated body must be an integral multiple of the elementary charge e is known as

  1. Electromagnetic induction
  2. Conservation of charge
  3. Polarisation of charge
  4. Quantisation of charge Correct answer.

Explanation

Charge is quantised: any isolated body's net charge is an integer multiple of the elementary charge e = 1.60 × 10⁻¹⁹ C, so fractional values such as 0.5e are never observed as a net charge. Conservation of charge is a separate principle about the total charge of an isolated system remaining constant, while polarisation and induction describe redistribution of charge, not its discrete values.

Question 22Electrostatics

A point charge of +2.0 µC is moved slowly from a point at potential 200 V to a point at potential 500 V. The work done by the external agent in moving the charge is

  1. 3.0 × 10⁻⁴ J
  2. 1.0 × 10⁻³ J
  3. 6.0 × 10⁻⁴ J Correct answer.
  4. 4.0 × 10⁻⁴ J

Explanation

The work done by an external agent equals q(V_final − V_initial), because moving the charge slowly means there is no change in kinetic energy. Here q = 2.0 × 10⁻⁶ C and the potential rise is 500 V − 200 V = 300 V, giving W = 2.0 × 10⁻⁶ × 300 = 6.0 × 10⁻⁴ J. The electric field itself does −6.0 × 10⁻⁴ J of work during this displacement.

Question 23Current Electricity

The graph of current I (y-axis) against potential difference V (x-axis) for a metallic conductor kept at constant temperature is

  1. a straight line with a positive intercept on the V-axis
  2. a straight line passing through the origin Correct answer.
  3. a parabola curving upwards
  4. a rectangular hyperbola

Explanation

For an ohmic conductor at constant temperature the current is directly proportional to the potential difference, since V = IR with R fixed. The I–V graph is therefore a straight line whose slope equals 1/R and which passes through the origin, because zero p.d. means zero current.

Question 24Current Electricity

The SI unit of the resistivity of a conducting material is

  1. ohm (Ω)
  2. ohm per metre (Ω m⁻¹)
  3. siemens per metre (S m⁻¹)
  4. ohm metre (Ω m) Correct answer.

Explanation

Resistivity ρ is defined by R = ρL/A, so ρ = RA/L and its unit is (Ω × m²)/m = Ω m. Resistivity is a property of the material itself and does not depend on the length or area of any particular sample.

Question 25Current Electricity

Three resistors of 2 Ω, 3 Ω and 5 Ω are joined in series. The equivalent resistance of the combination is

  1. 10 Ω Correct answer.
  2. 0.97 Ω
  3. 30 Ω
  4. 5 Ω

Explanation

In a series combination the same current passes through every resistor and the potential differences add, so the resistances simply add: R = 2 + 3 + 5 = 10 Ω. The equivalent resistance of a series network is always greater than the largest individual resistance.

Question 26Current Electricity

A metal wire has a resistance of 10.0 Ω at 20 °C and a temperature coefficient of resistance of 4.0 × 10⁻³ °C⁻¹. Its resistance at 120 °C is

  1. 14.0 Ω Correct answer.
  2. 12.0 Ω
  3. 10.4 Ω
  4. 40.0 Ω

Explanation

Over this range the resistance varies as R = R₀[1 + α(T − T₀)], with R₀ = 10.0 Ω and α = 4.0 × 10⁻³ °C⁻¹. Here the rise is ΔT = 120 − 20 = 100 °C, so R = 10.0 × (1 + 4.0 × 10⁻³ × 100) = 10.0 × 1.40 = 14.0 Ω.

Question 27Current Electricity

A copper wire of cross-sectional area 1.0 × 10⁻⁶ m² carries a current of 1.6 A. If the free-electron density is 8.5 × 10²⁸ m⁻³, the drift speed of the electrons is about

  1. 1.2 × 10⁻⁶ m s⁻¹
  2. 1.2 × 10⁻⁴ m s⁻¹ Correct answer.
  3. 1.2 × 10² m s⁻¹
  4. 1.2 × 10⁻² m s⁻¹

Explanation

The current is related to drift speed by I = nAev_d, so v_d = I/(nAe). Substituting 1.6 A, n = 8.5 × 10²⁸ m⁻³, A = 1.0 × 10⁻⁶ m² and e = 1.6 × 10⁻¹⁹ C gives nAe ≈ 1.36 × 10⁴, hence v_d ≈ 1.2 × 10⁻⁴ m s⁻¹, a typical very slow drift speed.

Question 28Current Electricity

A cell of emf 12 V and internal resistance 2.0 Ω is connected across a 10 Ω resistor. The terminal potential difference of the cell is

  1. 14.4 V
  2. 10 V Correct answer.
  3. 12 V
  4. 2.0 V

Explanation

The total resistance of the circuit is R + r = 10 + 2 = 12 Ω, so the current is I = ε/(R + r) = 12/12 = 1.0 A. The terminal potential difference is then V = ε − Ir = 12 − (1.0)(2.0) = 10 V, which is also IR across the external resistor.

Question 29Current Electricity

An electric bulb marked 100 W, 220 V is operated on a 110 V supply. Assuming its resistance stays constant, the power consumed by the bulb is

  1. 50 W
  2. 200 W
  3. 25 W Correct answer.
  4. 100 W

Explanation

The resistance is fixed by the rating, R = V²/P = 220²/100 = 484 Ω. At 110 V the power is V²/R = 110²/484 = 25 W, one quarter of the rated value, because at constant resistance the power is proportional to the square of the applied voltage.

Question 30Current Electricity

A potentiometer is preferred to an ordinary voltmeter for measuring the emf of a cell because, at the balance point,

  1. the cell under test draws no current Correct answer.
  2. the potentiometer wire carries no current
  3. the galvanometer shows a large steady deflection
  4. the potential drop along the whole wire becomes zero

Explanation

At the null point the potential difference across the balancing length of the wire exactly equals the emf of the cell and the galvanometer shows no deflection. Since no current is drawn from the cell, its terminal potential difference equals its emf, so the measurement is accurate; a voltmeter always draws some current and therefore reads slightly less than the emf.

Question 31Current Electricity

In a metre bridge experiment a resistance of 6 Ω is placed in the left gap and the balance point is found at 40 cm from the left end of the wire. The unknown resistance in the right gap is

  1. 15 Ω
  2. 9 Ω Correct answer.
  3. 4 Ω
  4. 6 Ω

Explanation

At balance the metre bridge satisfies R/S = l/(100 − l), where l is the balancing length measured from the end next to R. Hence S = R(100 − l)/l = 6 × 60/40 = 9 Ω, which is larger than the left-gap resistance because the balance point lies nearer the left end.

Question 32Current Electricity

A 60 W, 220 V lamp and a 100 W, 220 V lamp are connected in series across a 220 V supply. Which lamp glows more brightly?

  1. the 100 W lamp, because its resistance is lower
  2. the 60 W lamp, since its resistance is higher Correct answer.
  3. the 100 W lamp, because its rated power is greater
  4. both glow equally, since both are rated 220 V

Explanation

In a series circuit both lamps carry the same current, so the power actually dissipated in each is I²R. The 60 W lamp has the greater resistance, since R = V²/P gives 806.7 Ω against 484 Ω for the 100 W lamp, so it dissipates more power and glows more brightly.

Question 33Current Electricity

The commercial unit of electrical energy used in electricity bills is the kilowatt-hour. Expressed in joules, 1 kWh is equal to

  1. 3.6 × 10⁶ J Correct answer.
  2. 3.6 × 10³ J
  3. 1.6 × 10⁻¹⁹ J
  4. 9.0 × 10⁹ J

Explanation

One kilowatt-hour is the energy delivered at the rate of 1 000 W for 3 600 s, so 1 kWh = 1 000 × 3 600 = 3.6 × 10⁶ J. This is the unit in which domestic consumption is measured by the electricity meter and billed.

Question 34Current Electricity

Kirchhoff's junction rule states that the algebraic sum of the currents meeting at a node in a circuit is zero. This rule is a direct consequence of the conservation of

  1. energy
  2. mass
  3. momentum
  4. charge Correct answer.

Explanation

At a junction the total current flowing in equals the total current flowing out, so no net charge accumulates at the node. The junction rule is therefore a statement of conservation of charge, whereas the loop rule follows from conservation of energy.

Question 35Current Electricity

A carbon resistor is marked with the colour bands green, blue, red and gold in that order, starting from the end nearest a band. The resistance and tolerance of the resistor are

  1. 56 000 Ω ± 10 %
  2. 560 Ω ± 5 %
  3. 56 Ω ± 5 %
  4. 5 600 Ω ± 5 % Correct answer.

Explanation

Green stands for 5 and blue for 6, giving the significant figures 56; red is the multiplier 10², so the value is 56 × 100 = 5 600 Ω. The fourth band, gold, indicates a tolerance of ± 5 %.

Question 36Current Electricity

A uniform metallic wire of resistance R is stretched until its length becomes double, its volume remaining unchanged. The resistance of the stretched wire is

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

Explanation

Stretching at constant volume doubles the length and therefore halves the cross-sectional area. Since R = ρL/A, doubling L and halving A together multiply the resistance by four, giving 4R.

Question 37Current Electricity

A 4 Ω resistor and a 6 Ω resistor are joined in parallel across a battery of emf 12 V having negligible internal resistance. The current flowing through the 6 Ω resistor is

  1. 3.0 A
  2. 1.2 A
  3. 5.0 A
  4. 2.0 A Correct answer.

Explanation

In a parallel combination each branch has the full battery emf of 12 V across it. The current in the 6 Ω branch is therefore V/R = 12/6 = 2.0 A, quite independent of the 4 Ω branch.

Question 38Current Electricity

A steady potential difference V is maintained across a metallic wire of fixed length at constant temperature. If V is doubled, the drift velocity of the free electrons in the wire

  1. is halved
  2. becomes four times
  3. remains unchanged
  4. is doubled Correct answer.

Explanation

The drift velocity is given by v_d = eEτ/m, so it is proportional to the electric field inside the wire. For a wire of fixed length E = V/L, hence doubling V doubles both the field and the drift velocity, while the relaxation time τ stays the same at constant temperature.

Question 39Current Electricity

Two identical cells, each of emf 1.5 V and internal resistance 0.5 Ω, are connected in parallel and the combination is joined across a 1.25 Ω resistor. The current through the resistor is

  1. 1.2 A
  2. 0.67 A
  3. 0.86 A
  4. 1.0 A Correct answer.

Explanation

For two identical cells in parallel the equivalent emf is 1.5 V and the equivalent internal resistance is 0.5/2 = 0.25 Ω. The total resistance of the circuit is 1.25 + 0.25 = 1.50 Ω, so the current is 1.5/1.5 = 1.0 A.

Question 40Current Electricity

An electric heater rated 1500 W is used for 2 hours every day. The electrical energy consumed by it in 30 days, expressed in commercial units (kWh), is

  1. 90 kWh Correct answer.
  2. 30 kWh
  3. 180 kWh
  4. 45 kWh

Explanation

The power in kilowatts is 1.5 kW and the heater runs for 2 h each day. Energy = 1.5 × 2 × 30 = 90 kWh, that is, 90 commercial units, which is what the meter would record.

Question 41Current Electricity

A galvanometer of resistance 99 Ω gives full-scale deflection for a current of 1 mA. The shunt resistance required to convert it into an ammeter of range 10 mA is

  1. 99 Ω
  2. 9 Ω
  3. 10 Ω
  4. 11 Ω Correct answer.

Explanation

The shunt carries the excess current, so when 1 mA passes through the galvanometer the shunt must carry 9 mA at the same potential difference. Hence S = I_g G/(I − I_g) = (1 × 99)/9 = 11 Ω, and the ammeter then reads 10 mA at full scale.

Question 42Current Electricity

A battery of emf 12 V and internal resistance 2 Ω is connected to a variable external resistor R. The maximum power that can be delivered to the external resistor is

  1. 36 W
  2. 18 W Correct answer.
  3. 9 W
  4. 72 W

Explanation

The power in the external resistor is greatest when R is made equal to the internal resistance r. At that setting P = E²/(4r) = 144/(4 × 2) = 18 W, and half of the total power generated is wasted inside the battery.

Question 43Current Electricity

The drift speed of the free electrons in a copper wire carrying a steady current of about 1 A is of the order of

  1. 10⁻¹⁴ m/s
  2. 10⁻⁴ m/s Correct answer.
  3. 10³ m/s
  4. 10⁸ m/s

Explanation

In a metal the free electrons move randomly with thermal speeds of about 10⁵–10⁶ m/s, but the small drift velocity superimposed on this random motion is only of the order of a fraction of a millimetre per second, that is about 10⁻⁴ m/s. The relation I = nAev_d shows that for ordinary currents and wire diameters the drift speed must be very small because the free-electron density n is extremely large.

Question 44Current Electricity

A uniform wire of total resistance 12 Ω is bent into a closed circle. The resistance measured between two points on the circle that are diametrically opposite each other is

  1. 12 Ω
  2. 6 Ω
  3. 24 Ω
  4. 3 Ω Correct answer.

Explanation

The two points divide the circle into two semicircular arcs of equal length, so each arc has resistance 12 Ω ÷ 2 = 6 Ω. These two arcs provide two parallel paths between the same pair of points, so the equivalent resistance R satisfies 1/R = 1/6 + 1/6 = 1/3, giving R = 3 Ω.

Question 45Magnetic Effects of Current and Magnetism

In the expression for the magnetic force on a moving charge, F = qvB sin θ, the SI unit of the magnetic flux density B is:

  1. weber per metre (Wb m⁻¹)
  2. newton metre per ampere (N m A⁻¹)
  3. ampere per metre (A m⁻¹)
  4. newton per ampere metre (N A⁻¹ m⁻¹) Correct answer.

Explanation

Rearranging F = qvB sin θ gives B = F/(qv sin θ), so the unit is N/(C m s⁻¹). Since one coulomb per second is one ampere, this reduces to N A⁻¹ m⁻¹, which is defined as the tesla, the SI unit of flux density.

Question 46Magnetic Effects of Current and Magnetism

Which statement correctly describes magnetic field lines around a bar magnet?

  1. They intersect one another at the neutral points of the magnet.
  2. They form closed continuous loops, running from the south pole to the north pole inside the magnet. Correct answer.
  3. They begin on the north pole and terminate on the south pole, so they are discontinuous inside the magnet.
  4. Their spacing is uniform everywhere, showing that the field of a bar magnet is uniform in all regions.

Explanation

Magnetic field lines are continuous closed curves: outside the magnet they are drawn from north to south, and inside the material they continue from south to north so that no line has an endpoint. This is a direct consequence of the absence of magnetic monopoles.

Question 47Magnetic Effects of Current and Magnetism

A long straight horizontal wire carries a current directed towards geographic north. At a point vertically above the wire, the magnetic field produced by this current is directed towards:

  1. east Correct answer.
  2. south
  3. west
  4. vertically downwards

Explanation

By the right-hand grip rule the field circles the wire in a plane perpendicular to it; with the current towards the north, the field at a point above the wire points towards the east. The field is always perpendicular both to the current and to the line joining the point to the wire.

Question 48Magnetic Effects of Current and Magnetism

A proton enters a uniform magnetic field with a velocity making an angle of 30° with the field direction. The path followed by the proton inside the field is:

  1. a circle
  2. a helix Correct answer.
  3. a parabola
  4. a straight line

Explanation

The velocity component perpendicular to the field provides the centripetal force and makes the proton circle, while the component parallel to the field is unaffected because the magnetic force on it is zero. Superposing the two motions gives a helical path whose axis lies along the field.

Question 49Magnetic Effects of Current and Magnetism

Two long parallel straight wires are 0.20 m apart and carry currents of 5.0 A and 8.0 A in the same direction. The force per unit length on each wire is:

  1. 4.0 × 10⁻⁵ N m⁻¹, repulsive
  2. 4.0 × 10⁻⁵ N m⁻¹, attractive Correct answer.
  3. 2.0 × 10⁻⁵ N m⁻¹, attractive
  4. 8.0 × 10⁻⁵ N m⁻¹, attractive

Explanation

Using F/L = μ₀I₁I₂/(2πd) gives (2 × 10⁻⁷ × 5.0 × 8.0)/0.20 = 4.0 × 10⁻⁵ N m⁻¹. Currents flowing in the same direction attract each other, as shown by applying the right-hand rule to find the field of one wire at the position of the other.

Question 50Magnetic Effects of Current and Magnetism

A proton and an alpha particle are accelerated from rest through the same potential difference and then enter a uniform magnetic field at right angles to it. The ratio of the radius of the proton's path to that of the alpha particle is:

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

Explanation

For a charge accelerated through a potential difference V, the radius r = mv/(qB) becomes r = √(2mV/q)/B, so r ∝ √(m/q). The alpha particle has four times the proton mass but twice the charge, giving √(4/2) = √2 times the proton radius.