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CHSE Odisha Elective Physics study materials
CHSE Odisha Elective Physics study resources

CHSE Odisha Elective Physics practice questions

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Question 1Electrostatics and Capacitors

Two point charges of +4 μC and −4 μC are separated by 0.2 m in vacuum. The magnitude of the electrostatic force between them is closest to which value?

  1. 3.6 N Correct answer.
  2. 1.8 N
  3. 0.9 N
  4. 7.2 N

Explanation

Coulomb's law gives F = k|q1 q2|/r². With k = 9×10⁹ N m² C⁻², q1 q2 = 16×10⁻¹² C² and r² = 0.04 m², so F = 9×10⁹ × 16×10⁻¹²/0.04 = 3.6 N. The opposite signs only fix the direction (attraction), not the magnitude.

Question 2Electrostatics and Capacitors

The electric field inside a hollow conducting sphere carrying a net charge, at a point with no charge present in the cavity, is

  1. directed radially outward from the centre
  2. zero everywhere inside the cavity Correct answer.
  3. equal to the field just outside the surface
  4. inversely proportional to distance from the centre

Explanation

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

Question 3Electrostatics and Capacitors

A parallel-plate capacitor of capacitance C is charged to potential V and then disconnected from the battery. If the plate separation is doubled, the stored energy becomes

  1. C V² Correct answer.
  2. ¼C V²
  3. 2C V²
  4. ½C V²

Explanation

With the battery removed the charge Q stays fixed. Doubling the separation halves the capacitance (C = ε₀A/d), and the energy U = Q²/2C therefore doubles. The initial energy was ½CV², so the new energy is CV².

Question 4Electrostatics and Capacitors

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

  1. 0.1 mJ
  2. 100 μJ Correct answer.
  3. 1 μJ
  4. 25 μJ

Explanation

The work done by the external agent equals qΔV. Substituting q = 2×10⁻⁶ C and ΔV = 50 V gives W = 1×10⁻⁴ J = 100 μJ. The path taken between the points does not affect this result because the electrostatic field is conservative.

Question 5Electrostatics and Capacitors

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

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

Explanation

For series capacitors 1/C = 1/2 + 1/3 + 1/6 = 1 μF⁻¹, so C = 1 μF. The equivalent capacitance in series is always smaller than the smallest individual capacitor.

Question 6Electrostatics and Capacitors

A dielectric slab of relative permittivity 4 completely fills the space between the plates of a capacitor while the battery remains connected. The stored energy changes by a factor of

  1. ¼
  2. 4 Correct answer.
  3. ½
  4. 2

Explanation

With the battery connected the potential V is fixed, and the capacitance rises to 4C. Since U = ½CV², the energy becomes four times its original value. The dielectric polarises and the battery supplies the extra energy.

Question 7Electrostatics and Capacitors

The electric flux through a closed surface enclosing charges +2q and −5q is

  1. −3q/ε₀ Correct answer.
  2. zero
  3. +7q/ε₀
  4. −7q/ε₀

Explanation

Gauss's law states that the net flux through a closed surface equals the enclosed net charge divided by ε₀. The net enclosed charge is 2q − 5q = −3q, so the flux is −3q/ε₀. Charges outside the surface contribute nothing to the net flux.

Question 8Electrostatics and Capacitors

An electric dipole of moment p is placed in a uniform field E with its axis at 60° to the field. The magnitude of the torque on the dipole is

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

Explanation

The torque on a dipole is τ = pE sin θ, where θ is the angle between p and E. At 60°, sin 60° = √3/2, giving τ = pE√3/2. The torque is maximum at 90° and zero when the dipole is aligned with the field.

Question 9Electrostatics and Capacitors

The electric potential at a distance r from an isolated point charge Q is V. The potential at a distance 3r from the same charge is

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

Explanation

For a point charge the potential varies as 1/r, since V = kQ/r. Tripling the distance therefore reduces the potential to one-third of its former value. Potential, unlike field, falls off as the first power of distance.

Question 10Electrostatics and Capacitors

A 10 μF capacitor is charged to 100 V and then connected in parallel with an uncharged 10 μF capacitor. The common potential finally reached is

  1. 0 V
  2. 25 V
  3. 50 V Correct answer.
  4. 100 V

Explanation

Charge is conserved because the capacitors are isolated from any source: Q = 10 μF × 100 V = 1000 μC. The total capacitance becomes 20 μF, so the common potential is Q/C = 1000 μC / 20 μF = 50 V.

Question 11Electrostatics and Capacitors

The electric field on the axis of a uniformly charged ring of radius R at a distance x from its centre is maximum when x equals

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

Explanation

The axial field of a ring is E = kQx/(x² + R²)^{3/2}. Differentiating with respect to x and setting the derivative to zero gives x = R/√2. At this point the field is strongest along the axis.

Question 12Electrostatics and Capacitors

A charged particle enters a uniform electric field with its velocity parallel to the field lines. Its path inside the field is

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

Explanation

The force on the charge is qE, which is parallel to the velocity when the velocity is along the field. There is no transverse component, so the particle merely accelerates or decelerates along its original direction and travels in a straight line.

Question 13Electrostatics and Capacitors

The capacitance of an isolated conducting sphere of radius 0.1 m in vacuum is closest to

  1. 0.11 μF
  2. 9 pF
  3. 11 pF Correct answer.
  4. 1.1 nF

Explanation

For an isolated sphere C = 4πε₀R. With R = 0.1 m, C = 0.1/(9×10⁹) F ≈ 1.1×10⁻¹¹ F, that is about 11 pF. This small value shows why practical capacitors use closely spaced plates.

Question 14Electrostatics and Capacitors

The electric field just outside the surface of a charged conductor is 4×10⁴ N C⁻¹, directed normally outward. The surface charge density on the conductor is closest to

  1. 4.4×10⁻⁷ C m⁻²
  2. 3.5×10⁻⁷ C m⁻² Correct answer.
  3. 8.8×10⁻⁷ C m⁻²
  4. 1.1×10⁻⁷ C m⁻²

Explanation

At the surface of a conductor E = σ/ε₀, so σ = ε₀E. Using ε₀ = 8.85×10⁻¹² F m⁻¹ and E = 4×10⁴ N C⁻¹ gives σ ≈ 3.5×10⁻⁷ C m⁻². The field is normal to the surface because any tangential component would move the free charges.

Question 15Electrostatics and Capacitors

Two identical small conducting spheres carry charges +6 μC and −2 μC. They are touched together and then separated. The charge on each sphere is

  1. +4 μC
  2. +2 μC Correct answer.
  3. −4 μC
  4. +8 μC

Explanation

When identical conductors touch, the total charge is shared equally. The total is +6 μC + (−2 μC) = +4 μC, so each sphere ends with +2 μC. The final charges have the same sign because the net charge is positive.

Question 16Electrostatics and Capacitors

A parallel-plate capacitor has plates of area A and separation d. A dielectric slab of relative permittivity 2 and thickness d/2 is inserted, filling the space between one plate and the mid-plane. The new capacitance is

  1. (3/4)ε₀A/d
  2. (1/2)ε₀A/d
  3. (4/3)ε₀A/d Correct answer.
  4. 2ε₀A/d

Explanation

The arrangement is two capacitors in series: one of thickness d/2 with air, capacitance 2ε₀A/d, and one with dielectric, capacitance 4ε₀A/d. Their series combination gives 1/C = d/(2ε₀A) + d/(4ε₀A) = 3d/(4ε₀A), so C = (4/3)ε₀A/d.

Question 17Electrostatics and Capacitors

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

  1. kq₁q₂/r Correct answer.
  2. k(q₁+q₂)/r
  3. kq₁q₂r
  4. kq₁q₂/r²

Explanation

The potential energy of the pair is U = kq₁q₂/r, obtained by bringing one charge from infinity to distance r in the field of the other. It is negative for unlike charges and positive for like charges. Note that the field, not the energy, varies as 1/r².

Question 18Electrostatics and Capacitors

A Gaussian surface encloses a dipole consisting of charges +q and −q. The net electric flux through the surface is

  1. −q/ε₀
  2. 2q/ε₀
  3. zero Correct answer.
  4. q/ε₀

Explanation

Gauss's law depends only on the net enclosed charge, which for a dipole is +q + (−q) = 0. Hence the net flux through any closed surface enclosing the dipole is zero. Field lines enter and leave the surface in equal numbers.

Question 19Electrostatics and Capacitors

The electric field at a point on the perpendicular bisector of an electric dipole, far from the dipole compared with its length, varies with distance r as

  1. 1/r²
  2. 1/r
  3. 1/r⁴
  4. 1/r³ Correct answer.

Explanation

On the perpendicular bisector the fields of the two charges have a small resultant because their components along the axis cancel. The surviving transverse field falls off as p/(4πε₀r³), that is as 1/r³. This faster fall-off than 1/r² distinguishes a dipole from a single charge.

Question 20Electrostatics and Capacitors

A capacitor of 5 μF is charged to 200 V. The energy stored in it is

  1. 0.5 J
  2. 0.1 J Correct answer.
  3. 0.2 J
  4. 1 J

Explanation

The stored energy is U = ½CV². Substituting C = 5×10⁻⁶ F and V = 200 V gives U = ½ × 5×10⁻⁶ × 4×10⁴ = 0.1 J. The energy is held in the electric field between the plates.

Question 21Current Electricity

A copper wire of uniform cross-section carries a steady current. If the drift velocity of free electrons is v, what happens to the drift velocity when the current is doubled while the wire is kept at the same temperature?

  1. It becomes 2v Correct answer.
  2. It stays equal to v
  3. It becomes v/2
  4. It becomes 4v

Explanation

For a fixed conductor at constant temperature, I = nAev, so the drift velocity is directly proportional to the current because n, A and e are unchanged. Doubling the current therefore doubles the drift velocity. Note that the random thermal speed of electrons is far larger than the drift speed and is not the quantity meant here.

Question 22Current Electricity

Which statement correctly describes the resistivity of a metallic conductor?

  1. It increases when the conductor is cooled below room temperature
  2. It equals the resistance of a conductor of any length
  3. It depends on the dimensions of the specimen but not on its material
  4. It is independent of the specimen's dimensions and depends on the material and temperature Correct answer.

Explanation

Resistivity is an intrinsic property of the material, so it does not change when the length or area of a particular specimen is altered. For metals it rises with temperature because lattice vibrations scatter the electrons more strongly, and it falls on cooling. Resistance, not resistivity, depends on the geometry through R = ρL/A.

Question 23Current Electricity

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

  1. 11 Ω
  2. 1 Ω Correct answer.
  3. 6 Ω
  4. 0.5 Ω

Explanation

For a parallel combination, 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so R = 1 Ω. The equivalent resistance of a parallel network is always smaller than the smallest individual resistance, which here is 2 Ω, so the answer must be below 2 Ω.

Question 24Current Electricity

Kirchhoff's junction rule is a direct consequence of which conservation principle in a steady-current circuit?

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

Explanation

The junction rule states that the total current entering a node equals the total current leaving it, which expresses the fact that charge does not accumulate at a steady-state node. The loop rule, by contrast, follows from conservation of energy. Together these two rules are sufficient to analyse any resistive network.

Question 25Current Electricity

The terminals of a cell of emf 12 V and internal resistance 1 Ω are connected to an external resistance of 5 Ω. What is the terminal potential difference of the cell?

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

Explanation

The current is I = ε/(R + r) = 12/(5 + 1) = 2 A. The terminal potential difference is V = ε − Ir = 12 − (2)(1) = 10 V, which also equals IR = 2 × 5. The terminal voltage is always less than the emf while the cell delivers current.

Question 26Current Electricity

A wire of resistance R is stretched uniformly until its length is doubled. Assuming the resistivity and volume stay constant, what is its new resistance?

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

Explanation

Stretching keeps the volume constant, so doubling the length halves the cross-sectional area. Since R = ρL/A, the length factor of 2 and the area factor of 1/2 together multiply the resistance by 4. Hence the new resistance is 4R.

Question 27Current Electricity

Two cells of emf 6 V and 4 V with internal resistances 1 Ω and 2 Ω respectively are connected in parallel with their positive terminals together. What is the equivalent emf of the combination?

  1. 16/3 V Correct answer.
  2. 10 V
  3. 5 V
  4. 2 V

Explanation

For cells in parallel, the equivalent emf is ε = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) = (6×2 + 4×1)/(1 + 2) = 16/3 V. The equivalent internal resistance is r₁r₂/(r₁ + r₂) = 2/3 Ω. This formula applies when the cells send current in the same sense around the external circuit.

Question 28Current Electricity

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

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

Explanation

At balance the ratio of the resistances equals the ratio of the balancing lengths: X/6 = 40/60. This gives X = 6 × (40/60) = 4 Ω. The bridge works because the two gaps and the uniform wire form two parallel potential dividers whose midpoints are at the same potential at balance.

Question 29Current Electricity

A battery of emf 10 V and internal resistance 2 Ω delivers maximum power to an external resistor. What is the value of that maximum power?

  1. 5 W
  2. 12.5 W Correct answer.
  3. 50 W
  4. 25 W

Explanation

Maximum power transfer to an external load occurs when the load resistance equals the internal resistance, here 2 Ω. The current is then 10/(2 + 2) = 2.5 A, and the power in the load is I²R = (2.5)² × 2 = 12.5 W. This is the maximum power the source can deliver to any purely resistive load.

Question 30Current Electricity

The temperature coefficient of resistance of a metal is positive. What is the correct behaviour of a metallic filament as its temperature rises?

  1. Its resistance stays constant because the material is unchanged
  2. Its resistance rises only if the current is kept constant
  3. Its resistance falls because electrons move faster
  4. Its resistance rises because collisions with vibrating ions become more frequent Correct answer.

Explanation

In a metal, higher temperature increases the amplitude of lattice vibrations, so the mean free time between electron collisions falls. Since resistivity is proportional to 1/τ, both resistivity and resistance increase, giving a positive temperature coefficient. Faster electron motion does not increase the current because the random thermal velocity does not contribute to net drift.

Question 31Current Electricity

In the circuit, a 12 V battery is connected in series with 4 Ω and 8 Ω resistors. What is the potential difference across the 8 Ω resistor?

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

Explanation

The series current is 12/(4 + 8) = 1 A. The potential difference across the 8 Ω resistor is therefore 1 × 8 = 8 V, while the 4 Ω resistor takes 4 V; the two drops add to the battery emf. In a series circuit the larger resistance always takes the larger share of the applied voltage.

Question 32Current Electricity

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

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

Explanation

The shunt carries the excess current, so Ig G = (I − Ig) S. Here 1 mA × 50 Ω = 9 mA × S, giving S = 50/9 ≈ 5.56 Ω. A small shunt in parallel diverts most of the current and protects the coil, which is why ammeters have very low resistance.

Question 33Current Electricity

A galvanometer of resistance 100 Ω and full-scale current 1 mA is to be converted into a voltmeter of range 10 V. What series resistance is required?

  1. 10 000 Ω
  2. 9.9 Ω
  3. 9900 Ω Correct answer.
  4. 100 Ω

Explanation

For a voltmeter the multiplier resistance is in series: V = Ig(Rg + R). Thus 10 = 0.001(100 + R), so 100 + R = 10 000 and R = 9900 Ω. The high series resistance limits the current through the coil to its full-scale value at the rated voltage.

Question 34Current Electricity

Twelve identical cells, each of emf 2 V and internal resistance 0.5 Ω, are connected in series with an external resistance of 6 Ω. What is the current in the circuit?

  1. 4 A
  2. 2 A Correct answer.
  3. 24 A
  4. 1 A

Explanation

In series the emfs add and the internal resistances add: total emf = 12 × 2 = 24 V and total internal resistance = 12 × 0.5 = 6 Ω. The current is 24/(6 + 6) = 2 A. Grouping cells in series is useful only when the external resistance is much larger than the total internal resistance.

Question 35Current Electricity

A cell of emf ε and internal resistance r is connected to a variable external resistance R. Which graph correctly represents the terminal potential difference V against the current I drawn from the cell?

  1. A rectangular hyperbola
  2. A straight line of slope +r passing through the origin
  3. A straight line of slope −r with intercept ε on the V axis Correct answer.
  4. A parabola opening downward

Explanation

The terminal voltage obeys V = ε − Ir, which is linear in I with a negative slope equal to the internal resistance and an intercept equal to the emf. The intercept on the I axis is ε/r, the short-circuit current. This straight-line form is the basis of the standard method of finding ε and r from a V–I graph.

Question 36Current Electricity

A potential difference of 20 V is applied across a conductor of length 2 m and resistivity 1.6 × 10⁻⁸ Ω m, whose cross-sectional area is 2 × 10⁻⁶ m². What is the drift velocity of the conduction electrons, taking n = 8 × 10²⁸ m⁻³ and e = 1.6 × 10⁻¹⁹ C?

  1. 1.25 × 10⁻³ m/s Correct answer.
  2. 2.5 × 10⁻⁴ m/s
  3. 5 × 10⁻⁵ m/s
  4. 7.8 × 10⁻⁴ m/s

Explanation

First find the resistance: R = ρL/A = (1.6 × 10⁻⁸ × 2)/(2 × 10⁻⁶) = 1.6 × 10⁻² Ω. The current is I = V/R = 20/0.016 = 1250 A, and v = I/(nAe) = 1250/(8 × 10²⁸ × 2 × 10⁻⁶ × 1.6 × 10⁻¹⁹) ≈ 1.25 × 10⁻³ m/s. Drift speeds are small even for large currents because the free-electron density is enormous.

Question 37Magnetic Effects of Current and Magnetism

A long straight wire carries a steady current I. According to the Biot–Savart law, the magnetic field at a point in vacuum depends on the perpendicular distance r from the wire in which way?

  1. It is proportional to r
  2. It is proportional to 1/r Correct answer.
  3. It is independent of r
  4. It is proportional to 1/r²

Explanation

For an infinitely long straight conductor the field magnitude is B = μ₀I/(2πr), so it falls off as the first power of the inverse distance. The 1/r² dependence belongs to the field of a short current element or a point pole, not to a long wire.

Question 38Magnetic Effects of Current and Magnetism

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

  1. Zero, because the fields cancel midway
  2. Attractive only if one current is reversed
  3. Attractive, because the currents are parallel Correct answer.
  4. Repulsive, because the currents are parallel

Explanation

Each wire sits in the magnetic field of the other, and the Lorentz force F = IL × B on parallel currents points toward the other wire. Hence parallel currents attract, while antiparallel currents repel, with force per unit length μ₀I₁I₂/(2πd).

Question 39Magnetic Effects of Current and Magnetism

A charged particle enters a uniform magnetic field with its velocity parallel to the field lines. What path does it follow inside the field?

  1. A parabolic path that slows the particle
  2. A helical path along the field
  3. A circular path of constant radius
  4. A straight line with constant speed Correct answer.

Explanation

The magnetic force qv × B vanishes when v is parallel to B, since the cross product of parallel vectors is zero. With no magnetic force the particle continues in a straight line at constant speed, because magnetic forces never do work.

Question 40Magnetic Effects of Current and Magnetism

The magnetic field at the centre of a circular coil of N turns, radius r, carrying current I is given by:

  1. μ₀NIr/2
  2. μ₀NI/(4πr)
  3. μ₀I/(2πr)
  4. μ₀NI/(2r) Correct answer.

Explanation

Each turn contributes μ₀I/(2r) at the centre, and N identical turns add in the same direction, giving B = μ₀NI/(2r). This is the on-axis field formula evaluated at the centre of the loop.

Question 41Magnetic Effects of Current and Magnetism

A soft iron core is inserted into a solenoid carrying a steady current. The magnetic field inside the solenoid:

  1. Increases greatly because the iron becomes magnetised Correct answer.
  2. Decreases because iron blocks magnetic lines
  3. Reverses direction because of induced currents
  4. Stays the same because the current is unchanged

Explanation

Soft iron has a high relative permeability, so its domains align with the solenoid field and add a large magnetisation contribution. The result is B = μ₀μ_r n I, a field many times stronger than in air for the same current.

Question 42Magnetic Effects of Current and Magnetism

A galvanometer of resistance 5 Ω gives full-scale deflection at 1 mA. What shunt resistance converts it into an ammeter reading up to 1 A?

  1. About 0.5 Ω
  2. About 5000 Ω
  3. About 5 Ω
  4. About 0.005 Ω Correct answer.

Explanation

The shunt carries the excess current 1 A − 0.001 A ≈ 0.999 A while the coil drops 1 mA × 5 Ω = 5 mV. Hence S = 0.005/0.999 ≈ 0.005 Ω, a very small resistance placed in parallel.

Question 43Magnetic Effects of Current and Magnetism

A moving-coil galvanometer is converted into a voltmeter of range V by connecting a resistance R in series. If the galvanometer has resistance G and full-scale current I_g, then R equals:

  1. V/I_g + G
  2. V·I_g − G
  3. G − V/I_g
  4. V/I_g − G Correct answer.

Explanation

At full scale the total series resistance must be V/I_g so that current I_g flows. The multiplier resistance is therefore the total minus the coil resistance, R = V/I_g − G.

Question 44Magnetic Effects of Current and Magnetism

A bar magnet is dropped through a vertical copper tube. Compared with free fall in air, the magnet falls:

  1. Faster, because copper attracts the magnet
  2. More slowly, but only while entering the tube
  3. At the same rate, because copper is non-magnetic
  4. More slowly, because induced currents oppose the motion Correct answer.

Explanation

The changing flux induces eddy currents in the copper, and by Lenz's law these currents produce fields that oppose the magnet's motion. The retarding force reduces the acceleration below g throughout the fall, not just at entry.

Question 45Magnetic Effects of Current and Magnetism

A magnetic dipole of moment m is placed in a uniform field B at an angle θ to the field. The torque on it and its potential energy are respectively:

  1. mB sin θ and +mB cos θ
  2. mB cos θ and −mB sin θ
  3. mB and zero for all θ
  4. mB sin θ and −mB cos θ Correct answer.

Explanation

The torque is τ = m × B, whose magnitude is mB sin θ and which tends to align m with B. The potential energy U = −m·B = −mB cos θ is minimum at θ = 0, the stable aligned position.

Question 46Magnetic 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 varies with x as:

  1. B ∝ (R² + x²)^(−3/2) Correct answer.
  2. B ∝ (R² + x²)^(−1)
  3. B ∝ (R² + x²)^(3/2)
  4. B ∝ (R² + x²)^(−1/2)

Explanation

The on-axis field is B = μ₀NIR²/[2(R² + x²)^(3/2)], so the dependence is the inverse three-halves power. At x = 0 this reduces to μ₀NI/(2R) and it decreases as x grows.

Question 47Magnetic Effects of Current and Magnetism

An electron moves with speed v in a circle of radius r in a uniform magnetic field B. If the speed is doubled while B is unchanged, the radius and the time period become:

  1. Radius doubled, period unchanged Correct answer.
  2. Radius halved, period unchanged
  3. Radius doubled, period doubled
  4. Radius unchanged, period halved

Explanation

Setting qvB = mv²/r gives r = mv/(qB), so doubling v doubles r. The period T = 2πm/(qB) is independent of speed, so it stays the same.

Question 48Magnetic Effects of Current and Magnetism

A solenoid of length 0.5 m has 1000 turns and carries a current of 2 A in air. The field near its centre is approximately:

  1. 1.0 × 10⁻² T
  2. 2.5 × 10⁻³ T
  3. 4.0 × 10⁻⁶ T
  4. 5.0 × 10⁻³ T Correct answer.

Explanation

For a long solenoid B = μ₀nI with n = N/L = 1000/0.5 = 2000 turns per metre. Thus B = 4π × 10⁻⁷ × 2000 × 2 ≈ 5.0 × 10⁻³ T.

Question 49Magnetic Effects of Current and Magnetism

The horizontal component of the Earth's field at a place is 0.3 G and the angle of dip is 30°. The total field intensity there is:

  1. 0.60 G
  2. 0.15 G
  3. 0.35 G Correct answer.
  4. 0.26 G

Explanation

The horizontal component is H = B cos δ, so B = H/cos δ = 0.3/cos 30° = 0.3/0.866 ≈ 0.35 G. The vertical component would be B sin δ ≈ 0.17 G.

Question 50Magnetic Effects of Current and Magnetism

A rectangular loop of N turns, area A, carrying current I is free to rotate in a uniform field B, and its plane is initially parallel to B. Which statement about the initial torque is correct?

  1. It is maximum, equal to NIAB, because the normal is perpendicular to B Correct answer.
  2. It is maximum, equal to NIAB cos 45°
  3. It is zero, because the plane is parallel to B
  4. It is NIAB/2, because only two sides feel force

Explanation

Torque is τ = NIAB sin θ, where θ is measured between the loop normal and B. With the plane parallel to B the normal is at 90°, so sin θ = 1 and the torque is maximum, NIAB. All four sides contribute to the couple, though the forces on two sides are along the axis.