I85PREP
⌘ K
BSEB Inter Physics study materials
BSEB Inter Physics study resources

BSEB Inter Physics practice questions

One question at a time. Build your BSEB Inter 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 −2 μC are separated by 10 cm in air. What is the magnitude of the electrostatic force between them? (Take 1/(4πε₀) = 9 × 10⁹ N m² C⁻²)

  1. 0.36 N
  2. 1.8 N
  3. 36 N
  4. 3.6 N Correct answer.

Explanation

Use Coulomb's law F = k|q₁q₂|/r². Here k = 9 × 10⁹, |q₁q₂| = (2 × 10⁻⁶)(2 × 10⁻⁶) = 4 × 10⁻¹² C², and r² = (0.1)² = 0.01 m². Substituting gives F = (9 × 10⁹ × 4 × 10⁻¹²)/0.01 = 3.6 N.

Question 2Electrostatics

The electric field intensity at a point is defined as which of the following?

  1. The work done in bringing a unit positive charge from infinity to that point
  2. The force experienced by a unit negative charge placed at that point
  3. The force experienced by a unit positive test charge placed at that point Correct answer.
  4. The total charge enclosed by a surface around that point

Explanation

Electric field intensity is the force per unit positive charge, E = F/q₀, with direction along the force on a positive charge. The magnitude is measured in newton per coulomb. It is a vector quantity determined by the source charges, not by the test charge.

Question 3Electrostatics

An electric dipole is placed in a uniform external electric field with its axis parallel to the field. What is the net force and net torque on the dipole in this orientation?

  1. Net force is zero and net torque is zero Correct answer.
  2. Net force is maximum and net torque is zero
  3. Net force is maximum and net torque is maximum
  4. Net force is zero and net torque is maximum

Explanation

In a uniform field the forces on +q and −q are equal and opposite, so the net force is always zero. The torque is τ = pE sinθ, and when the dipole axis is parallel to the field θ = 0, so sinθ = 0 and the torque vanishes. This is the stable equilibrium orientation.

Question 4Electrostatics

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

Explanation

By Gauss's law the total flux through the closed cube surface is q/ε₀. Because the charge is at the centre, the flux is shared equally among the six faces by symmetry. Each face therefore carries q/(6ε₀).

Question 5Electrostatics

The work done in moving a charge along an equipotential surface is which of the following?

  1. Equal to the change in potential energy divided by the charge
  2. Maximum, because the field is strongest along the surface
  3. Zero, because there is no potential difference between any two points on the surface Correct answer.
  4. Equal to the product of the field and the path length

Explanation

Work done by the field is W = q(V₁ − V₂). On an equipotential surface every point has the same potential, so V₁ = V₂ and the work is zero. The field lines are perpendicular to such surfaces, so the field does no work along them.

Question 6Electrostatics

The capacitance of a parallel plate capacitor is C when the plates are separated by distance d in air. If the separation is doubled and a dielectric of constant K = 2 fills the space, what is the new capacitance?

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

Explanation

For a parallel plate capacitor C = Kε₀A/d. Doubling d halves the capacitance, while inserting a dielectric of K = 2 doubles it. The two changes cancel, so the new capacitance equals the original C.

Question 7Electrostatics

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

  1. 15 μF
  2. 2 μF
  3. 1.5 μF Correct answer.
  4. 0.67 μF

Explanation

For series capacitors 1/C = 1/C₁ + 1/C₂ + 1/C₃. Here 1/C = 1/3 + 1/6 + 1/6 = 2/3 μF⁻¹, so C = 3/2 = 1.5 μF. The equivalent is always smaller than the smallest individual capacitor.

Question 8Electrostatics

A parallel plate capacitor is charged and then disconnected from the battery. If the plates are now pulled farther apart, what happens to the potential difference between the plates?

  1. It decreases, because capacitance decreases
  2. It becomes zero, because the field disappears
  3. It increases, because charge stays constant while capacitance decreases Correct answer.
  4. It remains the same, because the charge is constant

Explanation

With the battery disconnected the charge Q is fixed. Increasing the separation reduces C = ε₀A/d, and since V = Q/C, a smaller capacitance gives a larger potential difference. The stored energy also rises because work is done in separating the plates.

Question 9Electrostatics

The electric potential at a distance r from an isolated point charge Q is given by which expression? (k = 1/(4πε₀))

  1. kQ/r Correct answer.
  2. kQ²/r
  3. kQr
  4. kQ/r²

Explanation

Electric potential is the work done per unit charge bringing a test charge from infinity to the point, giving V = kQ/r. It varies as 1/r, unlike the field which varies as 1/r². Potential is a scalar and may be positive or negative depending on the sign of Q.

Question 10Electrostatics

Two charges +4 μC and +1 μC are placed 30 cm apart in air. At what distance from the +4 μC charge, measured along the line joining them, is the electric field zero?

  1. 20 cm Correct answer.
  2. 24 cm
  3. 10 cm
  4. 15 cm

Explanation

The zero-field point lies between the charges, closer to the smaller one. Setting k(4)/x² = k(1)/(30 − x)² gives 2/x = 1/(30 − x), so 60 − 2x = x and x = 20 cm. At 20 cm from the larger charge the two fields are equal and opposite.

Question 11Electrostatics

A hollow spherical conductor carries a net charge on its outer surface. What is the electric field inside the cavity of the conductor?

  1. Equal to kQ/r² measured from the centre
  2. Directed radially outward and increasing toward the centre
  3. Zero everywhere inside the cavity Correct answer.
  4. Equal to the field at the outer surface

Explanation

For a conductor in electrostatic equilibrium the charge resides on the outer surface and the field within the conducting material is zero. A Gaussian surface enclosing the empty cavity contains no charge, so the flux and hence the field inside the cavity are zero.

Question 12Electrostatics

The electric potential energy of a system of two point charges q₁ and q₂ separated by distance r in vacuum is which of the following?

  1. k q₁q₂ r
  2. k q₁q₂ / r Correct answer.
  3. k (q₁ + q₂) / r
  4. k q₁q₂ / r²

Explanation

The potential energy equals the work needed to assemble the charges from infinity, giving U = k q₁q₂/r. It is negative for unlike charges and positive for like charges. It falls off as 1/r, in step with the potential of one charge seen by the other.

Question 13Electrostatics

A 2 μF capacitor is charged to a potential difference of 100 V. How much energy is stored in it?

  1. 0.1 J
  2. 0.02 J
  3. 1 J
  4. 0.01 J Correct answer.

Explanation

The stored energy is U = ½CV². Substituting C = 2 × 10⁻⁶ F and V = 100 V gives U = ½ × 2 × 10⁻⁶ × 10⁴ = 0.01 J. The factor of one half matters because the potential rises gradually as the capacitor charges.

Question 14Electrostatics

A charge is placed at rest in a uniform electric field. Which statement describes its subsequent motion?

  1. It remains at rest because the field exerts no force
  2. It accelerates along the field lines, and its path is a straight line Correct answer.
  3. It moves in a circle around the field direction
  4. It accelerates perpendicular to the field lines

Explanation

A charge in an electric field experiences a force F = qE along the field direction for a positive charge. Starting from rest, the acceleration is along that force, so the path is a straight line parallel to the field. No magnetic force acts here to curve the path.

Question 15Electrostatics

Two identical small conducting spheres carry charges +6 μC and −2 μC. They are brought into contact and then separated. What is the charge on each sphere?

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

Explanation

On contact the total charge is shared equally between identical spheres. The total is (+6) + (−2) = +4 μC, so each sphere receives +4/2 = +2 μC. Charge is conserved throughout the process.

Question 16Electrostatics

A dielectric slab of constant K is inserted to fill the space between the plates of an isolated charged capacitor. How do the electric field between the plates and the stored energy change?

  1. The field stays the same while the energy decreases by K
  2. Both the field and the stored energy increase by a factor K
  3. The field decreases by K while the energy increases by K
  4. Both the field and the stored energy decrease by a factor K Correct answer.

Explanation

For an isolated capacitor the free charge is fixed, and the dielectric reduces the field to E₀/K. The energy U = Q²/(2C) falls because C rises to KC, giving U₀/K. Both quantities therefore decrease by the factor K.

Question 17Current Electricity

A wire of resistance R is stretched uniformly until its length is doubled. Assuming the volume of the wire stays the same, what is its new resistance?

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

Explanation

Stretching keeps volume constant, so doubling the length halves the cross-sectional area. Resistance equals resistivity times length divided by area, so the length factor of 2 and the area factor of 1/2 multiply to give four times the original resistance.

Question 18Current Electricity

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

  1. It is the random speed of electrons between successive collisions.
  2. It increases when the wire is cooled while the current is kept constant.
  3. It is the average velocity gained by electrons in the direction opposite to the applied electric field. Correct answer.
  4. It equals the speed at which the electric field signal travels along the wire.

Explanation

Drift velocity is the small average velocity superimposed on random thermal motion, directed opposite to the field because electrons are negatively charged. It is of the order of a fraction of a millimetre per second, far smaller than both the random thermal speed and the signal speed.

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

Explanation

For a parallel combination the reciprocals add: 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1. The reciprocal of the equivalent resistance is therefore 1, so the equivalent resistance is 1 ohm.

Question 20Current Electricity

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

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

Explanation

The current is emf divided by total resistance, 12/6 = 2 A. The terminal voltage is emf minus the internal drop, 12 minus (2 times 2) = 8 V, which also equals the current times the external resistance.

Question 21Current Electricity

In the metre bridge experiment, a balance point is obtained at 40 cm from the left end when an unknown resistance is in the left gap and a 6 ohm standard resistor is in the right gap. What is the unknown resistance?

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

Explanation

At balance the unknown resistance divided by the standard equals the balancing length divided by the remaining length. Here that is 40/60, so the unknown is 6 times 2/3 = 4 ohm.

Question 22Current Electricity

What is the effect on the resistance of a metallic conductor when its temperature is raised, and why?

  1. Resistance increases because lattice vibrations impede electron drift more strongly. Correct answer.
  2. Resistance decreases because the relaxation time becomes longer.
  3. Resistance stays unchanged because the number of free electrons is fixed.
  4. Resistance decreases because electrons gain kinetic energy.

Explanation

In a metal, raising the temperature increases the amplitude of lattice vibrations, so electrons collide more frequently and the relaxation time shortens. Since resistivity is proportional to the inverse of the relaxation time, resistance rises with temperature.

Question 23Current Electricity

A battery of emf 6 V and negligible internal resistance is connected across a 3 ohm resistor. How much charge passes through the resistor in 2 minutes?

  1. 720 C
  2. 4 C
  3. 2 C
  4. 240 C Correct answer.

Explanation

The current is 6/3 = 2 A. Charge equals current times time, and 2 minutes is 120 s, so the charge is 2 times 120 = 240 C.

Question 24Current Electricity

Two cells of emf 2 V and 1.5 V with internal resistances 1 ohm and 0.5 ohm are connected in parallel, with like poles together, across a 3 ohm resistor. What is the current through the external resistor?

  1. 0.75 A
  2. 1.0 A
  3. 0.5 A
  4. 0.6 A Correct answer.

Explanation

The equivalent emf is (E1/r1 + E2/r2)/(1/r1 + 1/r2) = (2 + 3)/(1 + 2) = 5/3 V, and the equivalent internal resistance is 1/(1 + 2) = 1/3 ohm. Total resistance is 3 + 1/3 = 10/3 ohm, so the current is (5/3)/(10/3) = 0.5 A.

Question 25Current Electricity

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

  1. 1.2 ohm
  2. 2 ohm
  3. 0.5 ohm
  4. 1 ohm Correct answer.

Explanation

The internal resistance is R times (L1 - L2)/L2, where L1 is the open-circuit balance length and L2 the length with the resistor connected. That gives 5 times (60 - 50)/50 = 5 times 0.2 = 1 ohm.

Question 26Current Electricity

Which of these is the correct statement of Kirchhoff's junction rule?

  1. The sum of the potential differences around any closed loop is zero.
  2. The total emf in a circuit equals the total resistance multiplied by the current.
  3. The current through a conductor is proportional to the potential difference across it.
  4. The algebraic sum of the currents meeting at a junction is zero. Correct answer.

Explanation

The junction rule expresses conservation of charge: charge does not accumulate at a node, so the currents entering equal the currents leaving. The loop rule, by contrast, expresses conservation of energy around a closed path.

Question 27Current Electricity

A uniform wire of resistance 12 ohm is bent to form a closed circle. What is the resistance between two points that are diametrically opposite on the circle?

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

Explanation

Two diametrically opposite points divide the wire into two semicircular halves, each of resistance 6 ohm. These halves are in parallel between the chosen points, so the equivalent resistance is 6/2 = 3 ohm.

Question 28Current Electricity

A galvanometer of resistance 50 ohm gives full-scale deflection for a current of 2 mA. What shunt resistance converts it into an ammeter of range 1 A?

  1. 1.0 ohm
  2. 0.5 ohm
  3. 0.2 ohm
  4. 0.1 ohm Correct answer.

Explanation

The shunt carries the excess current, 1 - 0.002 = 0.998 A, while the galvanometer carries 0.002 A at full scale. Since the shunt and galvanometer are in parallel and share the same voltage, S = (0.002 times 50)/0.998, which is approximately 0.1 ohm.

Question 29Current Electricity

A heater of resistance 100 ohm carries a current of 2 A for 5 minutes. How much heat is produced?

  1. 24000 J
  2. 120000 J Correct answer.
  3. 1000 J
  4. 2000 J

Explanation

Joule heating is given by current squared times resistance times time. With 2 A, 100 ohm and 300 s, the heat is 4 times 100 times 300 = 120000 J.

Question 30Current Electricity

Two identical cells, each of emf E and internal resistance r, are connected in series with an external resistance R. For what value of R is the power delivered to the external resistor maximum?

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

Explanation

In series the two cells behave as a single source of emf 2E and internal resistance 2r. Maximum power transfer to the external load occurs when the load resistance equals the total internal resistance, so R = 2r.

Question 31Current Electricity

A copper wire and an iron wire of the same length and the same cross-sectional area are joined in series and a current is passed. Which quantity is the same for both wires?

  1. The number of free electrons per unit volume
  2. The current density Correct answer.
  3. The electric field inside the wire
  4. The potential difference per unit length

Explanation

In a series circuit the same current flows through both wires, and since their areas are equal the current density is identical. The electric field and the potential difference per unit length differ because they depend on the resistivity of each material.

Question 32Current Electricity

The resistivity of a semiconductor such as germanium behaves in which way as temperature rises?

  1. It increases steadily, as in a metal.
  2. It remains constant because the lattice is unaffected by heating.
  3. It first increases and then becomes zero at high temperature.
  4. It decreases, because the number of charge carriers increases rapidly. Correct answer.

Explanation

In a semiconductor, heating breaks more bonds and creates many additional electron-hole pairs, so the carrier concentration rises exponentially. This increase outweighs the reduced mobility, so the resistivity falls with temperature, unlike a metal.

Question 33Magnetic Effects of Current and Magnetism

A long straight wire carries a steady current I. At a perpendicular distance r from the wire, the magnitude of the magnetic field produced in vacuum is given by which expression?

  1. B = μ₀I/(2r²)
  2. B = μ₀I/(2πr) Correct answer.
  3. B = μ₀I/(4πr)
  4. B = μ₀Ir/(2π)

Explanation

Ampere's circuital law applied to a circular loop of radius r around a long straight conductor gives B(2πr) = μ₀I, so B = μ₀I/(2πr). The field falls off as 1/r and is tangential to circles centred on the wire. This is the standard result for an infinitely long straight current-carrying wire.

Question 34Magnetic Effects of Current and Magnetism

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

  1. Repulsive, because like currents repel
  2. Zero, because the fields cancel midway
  3. Attractive only if the currents are unequal
  4. Attractive, because the currents are parallel Correct answer.

Explanation

Each wire sits in the magnetic field of the other, and the Lorentz force on its charges points toward the neighbouring wire when the currents are parallel. Hence parallel currents attract and antiparallel currents repel, the opposite of the rule for like charges. The magnitude per unit length is μ₀I₁I₂/(2πd).

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 subsequent motion?

  1. Speed of the particle Correct answer.
  2. Magnitude of its acceleration
  3. Direction of its velocity
  4. Radius of its path

Explanation

The magnetic force qv×B is always perpendicular to velocity, so it does no work and the kinetic energy, hence the speed, stays constant. The velocity direction keeps turning and the acceleration vector rotates with it, while the radius stays fixed only because the speed is fixed. Speed is therefore the quantity that is genuinely conserved throughout.

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

Explanation

Each turn contributes μ₀I/(2R) at the centre, and N identical turns in series add their fields in the same direction. The result B = μ₀NI/(2R) shows the field grows with turn number and current and falls as the radius increases.

Question 37Magnetic Effects of Current and Magnetism

A magnetic dipole of moment m is placed in a uniform magnetic field B. The torque acting on it is given by:

  1. τ = m × B Correct answer.
  2. τ = mB sin θ directed along B
  3. τ = m · B
  4. τ = B × m divided by μ₀

Explanation

The torque on a current loop in a magnetic field is the vector product of the magnetic moment and the field, τ = m × B. Its magnitude is mB sin θ, so it is maximum when m is perpendicular to B and zero when they are aligned. The direction is perpendicular to both m and B.

Question 38Magnetic Effects of Current and Magnetism

A galvanometer of resistance 100 Ω gives full-scale deflection at 1 mA. What series resistance converts it into a voltmeter reading up to 10 V?

  1. 1000 Ω
  2. 9900 Ω Correct answer.
  3. 10 000 Ω
  4. 100 Ω

Explanation

For a voltmeter the multiplier resistance R is in series so that the full-scale current flows at the maximum voltage: V = I_g(R_g + R). Substituting gives 10 = 1×10⁻³(100 + R), so 100 + R = 10 000 and R = 9900 Ω.

Question 39Magnetic Effects of Current and Magnetism

A proton and an alpha particle enter a uniform magnetic field with the same speed, moving perpendicular to the field. The ratio of the radii of their circular paths (proton : alpha) is:

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

Explanation

The radius of the circular path is r = mv/(qB). For equal speeds, r is proportional to m/q, and the alpha particle has about four times the proton mass but twice the charge, giving m/q twice as large. Hence r_alpha = 2r_proton, so the ratio proton : alpha is 1 : 2.

Question 40Magnetic Effects of Current and Magnetism

Two long parallel wires 0.10 m apart carry currents of 5 A and 8 A in opposite directions. The magnitude of the force per unit length on either wire is approximately:

  1. 4 × 10⁻⁵ N/m
  2. 1.6 × 10⁻⁴ N/m
  3. 2 × 10⁻⁴ N/m
  4. 8 × 10⁻⁵ N/m Correct answer.

Explanation

The force per unit length is F/L = μ₀I₁I₂/(2πd). Substituting μ₀ = 4π×10⁻⁷ T·m/A, I₁ = 5 A, I₂ = 8 A and d = 0.10 m gives (2×10⁻⁷)(40)/(0.10) = 8×10⁻⁵ N/m. The currents being opposite only makes the force repulsive, not different in magnitude.

Question 41Magnetic Effects of Current and Magnetism

A circular coil of 50 turns and radius 0.10 m carries a current of 2 A. Its magnetic dipole moment is approximately:

  1. 0.314 A·m²
  2. 31.4 A·m²
  3. 6.28 A·m²
  4. 3.14 A·m² Correct answer.

Explanation

The magnetic moment of a coil is m = NIA, where A = πR² = π(0.10)² ≈ 0.0314 m². Thus m = 50 × 2 × 0.0314 ≈ 3.14 A·m². The moment is independent of the shape being circular beyond its area.

Question 42Magnetic Effects of Current and Magnetism

A charged particle moves parallel to a uniform magnetic field. What is the magnetic force on it?

  1. Zero, because the velocity and field are parallel Correct answer.
  2. qvB directed perpendicular to the field
  3. Maximum, because the full speed is available
  4. qvB directed along the field

Explanation

The magnetic force is F = qv×B, and the cross product vanishes when v is parallel (or antiparallel) to B. Hence no magnetic force acts, and the particle continues in a straight line with constant speed. The force is greatest when velocity is perpendicular to the field, not parallel.

Question 43Magnetic Effects of Current and Magnetism

A moving-coil galvanometer is converted into an ammeter by connecting a shunt resistance. Which statement about the conversion is correct?

  1. The shunt is in parallel and is large compared with the galvanometer resistance
  2. The shunt is in series and is small compared with the galvanometer resistance
  3. The shunt is in parallel and is small compared with the galvanometer resistance Correct answer.
  4. The shunt is in series and is large compared with the galvanometer resistance

Explanation

An ammeter must have very low resistance, so a low shunt is connected in parallel with the galvanometer to divert most of the current. Only a small fraction then passes through the coil, allowing large currents to be measured without damage. The shunt value is chosen so the coil still reaches full-scale deflection at the desired range.

Question 44Magnetic Effects of Current and Magnetism

A current-carrying loop lies in the plane of the page in a uniform magnetic field directed into the page. What is the net force and net torque on the loop?

  1. Net force is maximum and net torque is zero
  2. Net force is zero and net torque is maximum
  3. Net force is zero and net torque is zero Correct answer.
  4. Net force is maximum and net torque is maximum

Explanation

A uniform field exerts forces on opposite sides of the loop that are equal and opposite, so the net force is zero. Because the magnetic moment of the loop is perpendicular to the page, parallel or antiparallel to B, the angle θ is 0° or 180° and τ = mB sin θ vanishes. The loop is therefore in equilibrium in this orientation.

Question 45Magnetic Effects of Current and Magnetism

The magnetic field on the axis of a circular coil of radius R at a distance x from its centre is proportional to which factor?

  1. (R² + x²)^(-2)
  2. (R² + x²)^(-3/2) Correct answer.
  3. (R² + x²)^(-1/2)
  4. (R² + x²)^(-1)

Explanation

The on-axis field of a current loop is B = μ₀IR²/(2(R² + x²)^(3/2)), so the distance dependence enters as (R² + x²)^(-3/2). At x = 0 this reduces to μ₀I/(2R), the centre field, and far away it falls off as 1/x³ like a dipole.

Question 46Magnetic Effects of Current and Magnetism

A beam of electrons moving horizontally toward the east passes undeflected through crossed electric and magnetic fields. The electric field points vertically downward. In which direction does the magnetic field point?

  1. Vertically upward
  2. Toward the south
  3. Toward the north Correct answer.
  4. Toward the west

Explanation

For no deflection the electric and magnetic forces must cancel: qE = qv×B, so v×B must point opposite to E, i.e. upward for an electron. With v toward the east and the magnetic force upward, the right-hand rule for v×B gives B toward the north. The electron's negative charge is already accounted for by using the force direction.

Question 47Magnetic Effects of Current and Magnetism

A solenoid of length 0.50 m has 1000 turns and carries a current of 2 A. Treating it as long, the magnetic field near its centre is approximately:

  1. 2.5 × 10⁻³ T
  2. 5.0 × 10⁻³ T Correct answer.
  3. 1.0 × 10⁻² T
  4. 5.0 × 10⁻⁴ T

Explanation

For a long solenoid B = μ₀nI, where n = N/L = 1000/0.50 = 2000 turns per metre. Then B = (4π×10⁻⁷)(2000)(2) ≈ 5.0×10⁻³ T. The field is uniform near the centre and directed along the axis.

Question 48Magnetic Effects of Current and Magnetism

A rectangular coil of N turns, area A, carrying current I is free to rotate in a uniform field B. It is held with its plane parallel to the field and then released. Which statement about the motion is correct?

  1. The torque is maximum but the coil experiences no change in magnetic potential energy
  2. The torque is zero at release and the coil stays at rest
  3. The torque is maximum at release and the coil rotates toward the position where its plane is perpendicular to B Correct answer.
  4. The torque is maximum at release and the coil rotates toward the position where its plane is parallel to B

Explanation

With the plane parallel to B, the normal to the coil (the magnetic moment) is perpendicular to B, so τ = NIAB sin 90° = NIAB is maximum. The torque turns the coil so that its magnetic moment aligns with B, which corresponds to the plane becoming perpendicular to the field. In that final position the torque falls to zero.

Question 49Electromagnetic Induction and Alternating Current

A bar magnet is pushed with its north pole facing a coil. According to Lenz's law, the induced current in the coil produces a magnetic field that

  1. is directed parallel to the magnet's axis of motion
  2. assists the approach of the north pole
  3. opposes the approach of the north pole Correct answer.
  4. reverses direction continuously as the magnet moves

Explanation

Lenz's law states that the induced emf always drives a current whose magnetic effect opposes the change in flux that produced it. When a north pole approaches, the flux through the coil increases, so the induced field must act like a north pole facing the magnet to repel it. This opposition converts mechanical work into electrical energy.

Question 50Electromagnetic Induction and Alternating Current

The magnetic flux linked with a coil is given by Φ = 4t² + 2t (weber), where t is in seconds. The magnitude of the induced emf at t = 3 s is

  1. 14 V
  2. 38 V
  3. 26 V Correct answer.
  4. 44 V

Explanation

By Faraday's law the induced emf equals the rate of change of flux, so emf = dΦ/dt = 8t + 2. Substituting t = 3 s gives 8(3) + 2 = 26 V. The sign would only indicate direction, and the question asks for magnitude.