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Tamil Nadu HSE Second Year Physics study materials
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Tamil Nadu HSE Second Year Physics practice questions

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Question 1Electrostatics

In Gauss's law, the electric flux Φ = E A cos θ through a surface is expressed in which SI unit?

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

Explanation

Flux is the product of electric field (measured in N C⁻¹) and area (m²), so its SI unit is N m² C⁻¹, which is equivalently volt metre. The unit V m⁻¹ belongs to electric field itself, and C m⁻² to surface charge density.

Question 2Electrostatics

Two point charges are held a distance r apart in vacuum. If the separation alone is doubled, the magnitude of the Coulomb force between them becomes

  1. one half of the original value
  2. four times the original value
  3. one quarter of the original value Correct answer.
  4. twice the original value

Explanation

Coulomb's law gives F proportional to 1/r², so the force falls off as the inverse square of the separation when the charges and the medium are unchanged. Doubling r multiplies the force by 1/2², that is by one quarter.

Question 3Electrostatics

A hollow metallic sphere carrying a net charge is in electrostatic equilibrium. At points inside the solid metal of the sphere, the electric field is

  1. equal to the field just outside its surface
  2. zero Correct answer.
  3. perpendicular to the surface and non-zero
  4. directly proportional to the distance from the centre

Explanation

In electrostatic equilibrium the free electrons of a conductor rearrange until the field inside the metal vanishes, and any excess charge resides on the outer surface. A non-zero internal field would continue to drive charges and so contradict the state of equilibrium.

Question 4Electrostatics

An electric dipole of moment p is placed in a uniform electric field E, with its axis making an angle of 30° with the field direction. The magnitude of the torque acting on the dipole is

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

Explanation

The torque on a dipole in a uniform field has magnitude τ = pE sin θ, where θ is the angle between p and E. Since sin 30° equals 0.5, the torque here is pE/2; it reaches the maximum value pE only at 90°.

Question 5Electrostatics

A point charge of 8.85 × 10⁻¹² C is placed at the centre of a spherical Gaussian surface of radius 0.50 m. Taking ε₀ = 8.85 × 10⁻¹² F m⁻¹, the electric flux through the surface is

  1. 0.50 N m² C⁻¹
  2. 2 N m² C⁻¹
  3. 8.85 N m² C⁻¹
  4. 1 N m² C⁻¹ Correct answer.

Explanation

Gauss's law states that the flux through any closed surface equals q/ε₀, the charge enclosed divided by the permittivity of free space. Here q/ε₀ equals 1, and the radius of the surface does not enter, because the same charge is enclosed whatever the size of the surface.

Question 6Electrostatics

The electric potential at a distance r from a point charge Q is V. The potential at a distance 2r from a different point charge 4Q is

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

Explanation

For a point charge the potential is V = kQ/r, so it is proportional to the charge and inversely proportional to the distance. Changing Q to 4Q multiplies the potential by four, and doubling the distance halves the result, giving 2V.

Question 7Electrostatics

Three identical point charges, each of magnitude q, are fixed at the corners of an equilateral triangle of side a in vacuum. The total electrostatic potential energy of the system is, with k = 1/4πε₀,

  1. (3/2)kq²/a
  2. kq²/a
  3. 3kq²/a Correct answer.
  4. 6kq²/a

Explanation

The system contains exactly three interacting pairs of charges, each pair separated by a side a and contributing kq²/a to the energy. Adding the three equal contributions gives 3kq²/a, each pair being counted once only.

Question 8Electrostatics

A positive charge is moved slowly in an electrostatic field from a point at potential 20 V to a point at potential 5 V. The work done by the electric field per unit charge is

  1. 15 J C⁻¹ Correct answer.
  2. 20 J C⁻¹
  3. 5 J C⁻¹
  4. −15 J C⁻¹

Explanation

The field does work q(V_initial − V_final) on a positive charge that moves from a higher to a lower potential. Per unit charge this is 20 − 5 = 15 J C⁻¹, and the work is positive because the field itself pushes the charge towards the lower potential.

Question 9Electrostatics

A parallel plate capacitor is charged to a potential difference V and is then isolated by disconnecting the battery. A dielectric slab of dielectric constant K is inserted so that it completely fills the gap. If the original stored energy was U, the energy now stored is

  1. U/K²
  2. KU
  3. U
  4. U/K Correct answer.

Explanation

With the battery disconnected the charge Q on the plates cannot change, while the capacitance rises from C to KC. The stored energy Q²/2C therefore falls to U/K, the difference appearing as the work done in drawing the slab into the field.

Question 10Electrostatics

An electron initially at rest is accelerated through a potential difference of 100 V. Taking e = 1.6 × 10⁻¹⁹ C and m = 9.1 × 10⁻³¹ kg, its final speed is about

  1. 8.4 × 10⁷ m s⁻¹
  2. 5.9 × 10⁶ m s⁻¹ Correct answer.
  3. 2.9 × 10⁶ m s⁻¹
  4. 1.2 × 10⁶ m s⁻¹

Explanation

The work done on the electron by the field, eV, appears as kinetic energy ½mv², so v = √(2eV/m). Substitution gives √(3.2 × 10⁻¹⁷ / 9.1 × 10⁻³¹), which is about 5.9 × 10⁶ m s⁻¹, comfortably below the speed of light.

Question 11Electrostatics

A negatively charged body carries a net charge of –4.8 × 10⁻¹⁹ C. Taking the elementary charge as 1.6 × 10⁻¹⁹ C, the number of excess electrons on the body is

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

Explanation

Electric charge is quantised, so every charge is an integer multiple of e and can be written q = ne. Here n = 4.8 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 3, and the negative sign simply shows that the body has an excess of electrons rather than a deficit.

Question 12Electrostatics

A charge is carried slowly from one point to another on the same equipotential surface. The work done by the electrostatic field during this movement is

  1. qV
  2. zero Correct answer.
  3. infinite
  4. qV²

Explanation

Every point on an equipotential surface is at the same potential, so the potential difference between the starting and finishing points is zero. The work done by the field is W = qΔV = 0, which is why no work is needed to move a charge along such a surface.

Question 13Electrostatics

Two point charges fixed a certain distance apart in vacuum repel each other with a force F. If the space around them is filled with a medium of dielectric constant K, the separation remaining unchanged, the force between the charges becomes

  1. K F
  2. F
  3. F / K Correct answer.
  4. F / K²

Explanation

In a material medium the Coulomb force is reduced to F_medium = F_vacuum / K because the molecules of the medium polarise and partly screen the field of the charges. Since the charges and their separation are unchanged, only the factor 1/K appears.

Question 14Electrostatics

At a point in the field of a point charge, the electric potential is 20 V and the magnitude of the electric field is 40 V m⁻¹. The distance of this point from the charge is

  1. 0.8 m
  2. 0.05 m
  3. 0.5 m Correct answer.
  4. 2.0 m

Explanation

For the field of a point charge, V = kQ/r and E = kQ/r², so dividing one by the other gives r = V/E. Substituting the data, r = 20/40 = 0.5 m, so the point lies half a metre from the charge.

Question 15Electrostatics

A parallel plate capacitor in air has capacitance C. If the area of each plate is doubled and the plate separation is halved, the capacitance becomes

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

Explanation

For a parallel plate capacitor C = ε₀A/d, so capacitance is directly proportional to plate area and inversely proportional to separation. Doubling A doubles C and halving d doubles it once more, giving 2 × 2 × C = 4C.

Question 16Electrostatics

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

Explanation

Gauss's law gives the total flux through the closed cube as q/ε₀, independent of the size of the cube. Because the charge sits at the centre, the six faces are symmetrically placed, so each face receives one sixth of the total flux, namely q/(6ε₀).

Question 17Electrostatics

Charges of +4 μC and –4 μC are placed 0.20 m apart in air. The electric potential at the midpoint of the line joining them is

  1. 3.6 × 10⁵ V
  2. 1.8 × 10⁵ V
  3. 7.2 × 10⁵ V
  4. zero Correct answer.

Explanation

Electric potential is a scalar, so the potentials due to the two charges add algebraically with their signs. At the midpoint each charge is 0.10 m away, giving +k(4 μC)/0.10 and –k(4 μC)/0.10, and these equal and opposite terms cancel to zero.

Question 18Electrostatics

A capacitor of capacitance 100 pF is charged to a potential difference of 100 V. The energy stored in the capacitor is

  1. 5.0 × 10⁻⁶ J
  2. 1.0 × 10⁻⁶ J
  3. 5.0 × 10⁻⁷ J Correct answer.
  4. 1.0 × 10⁻⁸ J

Explanation

The energy stored in a charged capacitor is U = ½CV². With C = 100 pF = 1.0 × 10⁻¹⁰ F and V = 100 V, U = ½ × 1.0 × 10⁻¹⁰ × (100)² = 5.0 × 10⁻⁷ J.

Question 19Electrostatics

An infinitely long cylinder of radius R carries a uniform volume charge density ρ. Using Gauss's law, the magnitude of the electric field at a point inside the cylinder at distance r from its axis (r < R) is

  1. ρr/(2ε₀) Correct answer.
  2. ρ/(2ε₀r)
  3. ρR²/(2ε₀r)
  4. ρr/ε₀

Explanation

Choose a coaxial cylindrical Gaussian surface of radius r and length L inside the cylinder. The charge enclosed is ρ(πr²L) and the flux is E(2πrL), so equating E(2πrL) = ρπr²L/ε₀ gives E = ρr/(2ε₀). Thus the field grows linearly with r inside the charge distribution.

Question 20Electrostatics

A capacitor of capacitance C is charged to a potential V and then connected in parallel with an uncharged capacitor of capacitance 2C, the battery having been removed. The common potential and the loss of energy in the process are

  1. V/3 and CV²/3 Correct answer.
  2. V/3 and CV²/6
  3. V/2 and CV²/4
  4. 2V/3 and CV²/3

Explanation

Charge is conserved because the battery is disconnected: CV = (C + 2C)V′, so the common potential is V′ = V/3. The initial energy is ½CV² and the final energy is ½(3C)(V/3)² = CV²/6, so the loss is ½CV² – CV²/6 = CV²/3.

Question 21Electrostatics

Taking the SI unit, the electric dipole moment of a pair of charges +q and –q separated by a small distance 2a is expressed in

  1. coulomb metre (C m) Correct answer.
  2. coulomb per metre (C m⁻¹)
  3. volt per metre (V m⁻¹)
  4. newton metre per coulomb (N m C⁻¹)

Explanation

The dipole moment is defined as p = q(2a), where q is in coulomb and 2a is the separation in metre. Hence its SI unit is coulomb metre, C m, and its direction is taken from the negative charge to the positive charge.

Question 22Electrostatics

Two capacitors of capacitance 3 μF and 6 μF are connected in series across a potential difference of 12 V. The charge on each capacitor is

  1. 4 μC
  2. 24 μC Correct answer.
  3. 12 μC
  4. 72 μC

Explanation

In a series combination, both capacitors carry the same charge Q. The equivalent capacitance is C_s = (3 × 6)/(3 + 6) = 2 μF, so Q = C_s V = 2 μF × 12 V = 24 μC.

Question 23Electrostatics

A thin spherical shell of radius R carries a uniform surface charge density σ. Using Gauss's law, the electric field at a point just outside the shell is

  1. σ/ε₀ directed radially outward Correct answer.
  2. σR/ε₀ directed radially outward
  3. zero
  4. σ/(2ε₀) directed radially outward

Explanation

For a thin spherical shell, a Gaussian sphere just outside radius R encloses the total charge Q = 4πR²σ. Gauss's law gives E(4πR²) = Q/ε₀, hence E = σ/ε₀, directed radially outward for positive σ.

Question 24Electrostatics

An air-filled parallel plate capacitor has plate area A and plate separation d, giving capacitance C = ε₀A/d. A dielectric slab of thickness t (t < d) and dielectric constant K is placed between the plates, parallel to them. The new capacitance is

  1. Kε₀A / (d − t + t/K)
  2. ε₀A / (d − t + Kt)
  3. ε₀A / (d + t(K − 1))
  4. ε₀A / [d − t + (t/K)] Correct answer.

Explanation

The dielectric slab and the remaining air gap act as two capacitors in series. The air part has thickness d − t, while the dielectric part contributes an effective thickness t/K. Therefore the new capacitance is C' = ε₀A/[d − t + t/K].

Question 25Current Electricity

A metallic wire of cross-sectional area A carries a steady current I. If n is the free-electron density and e the magnitude of the electronic charge, the drift speed of the electrons is

  1. nAe/I
  2. InAe
  3. nAeI
  4. I/(nAe) Correct answer.

Explanation

In a conductor the current is given by I = nAev_d, where n is the free-electron density, A the cross-sectional area and e the electronic charge. Rearranging this relation gives v_d = I/(nAe), so for the same current a thicker wire has a smaller drift speed.

Question 26Current Electricity

The resistivity of a metallic conductor depends on its

  1. applied potential difference
  2. area of cross-section
  3. length
  4. temperature Correct answer.

Explanation

Resistivity is a property of the material itself, and for a metal it changes with temperature because the rate of collisions of the free electrons changes. Length and area of cross-section alter the resistance through R = ρl/A but leave the resistivity ρ unchanged.

Question 27Current Electricity

A carbon resistor carries the colour bands brown, black and red in that order. Its resistance is

  1. 10 Ω
  2. 10 kΩ
  3. 1 kΩ Correct answer.
  4. 100 Ω

Explanation

In the colour code brown = 1 and black = 0 give the two significant digits, while red = 2 is the multiplier 10². The value is therefore 10 × 10² Ω = 1000 Ω, that is 1 kΩ.

Question 28Current Electricity

A cell of emf 6 V and internal resistance 1 Ω is connected across a 5 Ω resistor. The terminal potential difference of the cell is

  1. 1 V
  2. 4 V
  3. 5 V Correct answer.
  4. 6 V

Explanation

The current in the circuit is I = ε/(R + r) = 6/(5 + 1) = 1 A. The terminal potential difference is V = ε − Ir = 6 − (1)(1) = 5 V, which agrees with V = IR = (1)(5) = 5 V.

Question 29Current Electricity

In a Wheatstone bridge the arms P and Q lie on one side of the galvanometer and the arms R and S on the other. If P = 4 Ω, Q = 6 Ω and R = 8 Ω when the bridge is balanced, the unknown resistance S is

  1. 18 Ω
  2. 5.3 Ω
  3. 3 Ω
  4. 12 Ω Correct answer.

Explanation

At the balance point the ratio P/Q equals R/S and the galvanometer carries no current. Rearranging gives S = QR/P = (6 × 8)/4 = 12 Ω.

Question 30Current Electricity

A potentiometer is preferred to a voltmeter for measuring the emf of a cell because

  1. its resistance is lower than that of a voltmeter
  2. it works without any driving cell in the circuit
  3. at balance it draws no current from the cell under test Correct answer.
  4. it can measure emfs of the order of kilovolts directly

Explanation

A voltmeter always draws some current from the cell, so it reads the terminal voltage rather than the emf. At the balance point of a potentiometer no current is drawn from the cell being tested, so the balance length gives the emf directly.

Question 31Current Electricity

Resistors of 4 Ω and 6 Ω are joined in series across a battery. The ratio of the power dissipated in the 4 Ω resistor to that in the 6 Ω resistor is

  1. 1 : 1
  2. 3 : 2
  3. 2 : 3 Correct answer.
  4. 9 : 4

Explanation

In a series circuit the same current flows through both resistors, so the power P = I²R is directly proportional to the resistance. Hence P₄ : P₆ = 4 : 6 = 2 : 3.

Question 32Current Electricity

Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel. The equivalent resistance of the combination is

  1. 1 Ω Correct answer.
  2. 3.7 Ω
  3. 11 Ω
  4. 0.5 Ω

Explanation

For resistors in parallel, 1/R = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1, so R = 1 Ω. The equivalent resistance is smaller than the smallest individual resistor, as it must be for a parallel combination.

Question 33Current Electricity

Two copper wires of the same material have lengths in the ratio 1 : 2 and radii in the ratio 2 : 1. The ratio of their resistances R₁ : R₂ is

  1. 1 : 2
  2. 8 : 1
  3. 1 : 4
  4. 1 : 8 Correct answer.

Explanation

For a wire of length l and radius r, R = ρl/(πr²), so R is proportional to l/r². Therefore R₁/R₂ = (l₁/l₂)(r₂/r₁)² = (1/2)(1/2)² = 1/8, giving the ratio 1 : 8.

Question 34Current Electricity

In a potentiometer experiment the balance length for a cell on open circuit is 60 cm. When the cell is shunted by a 10 Ω resistor the balance length falls to 50 cm. The internal resistance of the cell is

  1. 2 Ω Correct answer.
  2. 1.2 Ω
  3. 5 Ω
  4. 12 Ω

Explanation

The open-circuit balance length of 60 cm corresponds to the emf ε and the length of 50 cm to the terminal voltage V when the 10 Ω resistor is connected. Since V = εR/(R + r), the internal resistance is r = R(l₁ − l₂)/l₂ = 10 × (60 − 50)/50 = 2 Ω.

Question 35Current Electricity

The current density at a point inside a conductor is defined as the current per unit area of cross-section. Its SI unit is

  1. A m²
  2. A m⁻² Correct answer.
  3. A m⁻¹
  4. A m

Explanation

Current density J is obtained by dividing current by area, J = I/A. Since current is measured in amperes and area in square metres, the SI unit of current density is ampere per square metre, written as A m⁻².

Question 36Current Electricity

The reciprocal of the resistance of a conductor is called its

  1. reluctance
  2. conductivity
  3. conductance Correct answer.
  4. resistivity

Explanation

The reciprocal of resistance R is conductance G, so G = 1/R, and it is measured in siemens. Conductivity is a different quantity: it is the reciprocal of resistivity, not of resistance.

Question 37Current Electricity

For a conductor that obeys Ohm's law at constant temperature, if the potential difference across it is doubled, the current through it

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

Explanation

Ohm's law gives V = IR, so at constant resistance the current is directly proportional to the potential difference. Doubling V therefore doubles I, provided the temperature and resistance remain unchanged.

Question 38Current Electricity

The current drawn from a cell of emf E and internal resistance r is largest when the external resistance connected across it is

  1. equal to r
  2. much greater than r
  3. zero Correct answer.
  4. infinite

Explanation

The current is I = E/(R + r). For fixed E and r, the current increases as the external resistance R decreases, and it reaches its maximum value E/r when R = 0, that is, when the cell is short-circuited.

Question 39Current Electricity

A galvanometer of resistance G gives full-scale deflection for a current Ig. It is converted into an ammeter by connecting a shunt S across it. The maximum current that the ammeter can measure is

  1. Ig(1 + S/G)
  2. Ig(1 + G/S) Correct answer.
  3. Ig G/S
  4. Ig

Explanation

At full-scale deflection, the shunt carries a current IgG/S because the same potential difference appears across G and S. The total ammeter current is therefore Ig + IgG/S, which equals Ig(1 + G/S).

Question 40Current Electricity

In a metre bridge experiment, the balance point is obtained on the wire. The bridge is most sensitive when the balance point is located

  1. at either end of the wire
  2. at one-fourth of the wire from one end
  3. near the middle of the wire Correct answer.
  4. at three-fourths of the wire from one end

Explanation

A metre bridge is a form of Wheatstone bridge, and its sensitivity is greatest when the four arms have comparable resistances. This condition is approached when the balance point lies near the middle of the wire, giving ratios close to one.

Question 41Current Electricity

The temperature coefficient of resistance of a semiconductor is

  1. negative Correct answer.
  2. infinite
  3. positive
  4. zero

Explanation

In a semiconductor, raising the temperature increases the number of free charge carriers, so the resistance falls. Because resistance decreases with increasing temperature, the temperature coefficient of resistance is negative.

Question 42Current Electricity

A battery of emf E and internal resistance r is connected to a variable external resistance R. The power delivered to R is maximum when R is

  1. infinite
  2. zero
  3. r Correct answer.
  4. 2r

Explanation

The power in the external resistor is P = E²R/(R + r)². This expression is maximum when the external resistance equals the internal resistance, R = r, which is the maximum power transfer condition.

Question 43Current Electricity

Three identical cells, each of emf 2 V and internal resistance 0.5 Ω, are joined in series. If one of the cells is reversed, the net emf of the combination is

  1. 4 V
  2. 6 V
  3. 1.5 V
  4. 2 V Correct answer.

Explanation

With all three cells aiding, the total emf would be 2 + 2 + 2 = 6 V. Reversing one cell subtracts its emf, so the net emf becomes 2 + 2 − 2 = 2 V, while the internal resistances still add in series.

Question 44Current Electricity

A uniform metallic wire of resistance R is stretched uniformly until its length becomes twice the original length. Assuming its volume remains constant, its new resistance is

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

Explanation

When the wire is stretched, its volume remains constant, so doubling the length halves the area of cross-section. Since R = ρl/A, doubling l and halving A multiplies the resistance by 2 × 2 = 4, giving 4R.

Question 45Current Electricity

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

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

Explanation

In the steady state no electric charge can pile up at a junction, so the total current entering must equal the total current leaving. The junction rule is therefore simply a statement of the conservation of electric charge.

Question 46Current Electricity

In a conductor the free electrons drift with a speed of 3.0 × 10⁻⁴ m s⁻¹ when the electric field inside the conductor is 6.0 × 10⁻² V m⁻¹. The mobility of these electrons is

  1. 1.8 × 10⁻⁵ m² V⁻¹ s⁻¹
  2. 5.0 × 10⁻³ m² V⁻¹ s⁻¹ Correct answer.
  3. 5.0 × 10⁻³ m s⁻¹
  4. 2.0 × 10² m² V⁻¹ s⁻¹

Explanation

Mobility μ is defined as the drift velocity acquired per unit electric field, μ = v_d/E. Substituting the values gives μ = (3.0 × 10⁻⁴)/(6.0 × 10⁻²) = 5.0 × 10⁻³ m² V⁻¹ s⁻¹.

Question 47Current Electricity

Four identical cells, each of emf 2 V and internal resistance 0.4 Ω, are connected in parallel and the combination is joined across a 1.9 Ω resistor. The current through the resistor is

  1. 0.87 A
  2. 1.00 A Correct answer.
  3. 0.57 A
  4. 1.05 A

Explanation

For identical cells in parallel the equivalent emf equals the emf of one cell, 2 V, while the equivalent internal resistance is r/n = 0.4/4 = 0.1 Ω. The current is therefore I = 2/(1.9 + 0.1) = 1.00 A.

Question 48Current Electricity

A galvanometer of resistance 50 Ω shows full-scale deflection for a current of 1.0 mA. It is to be converted into a voltmeter that reads up to 10 V. The resistance to be connected in series with the galvanometer is

  1. 9 500 Ω
  2. 10 000 Ω
  3. 50 Ω
  4. 9 950 Ω Correct answer.

Explanation

In a voltmeter the multiplier is in series with the galvanometer, so at full-scale deflection V = I_g(G + R). Hence R = V/I_g − G = 10/(1.0 × 10⁻³) − 50 = 9 950 Ω.

Question 49Magnetism and Magnetic Effects of Electric Current

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

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

Explanation

Each element of the loop contributes according to the Biot–Savart law, and integrating around all N turns gives B = μ₀NI/2r at the centre. The field is directed perpendicular to the plane of the coil, with its sense fixed by the right-hand rule.

Question 50Magnetism and Magnetic Effects of Electric Current

The SI unit of magnetic dipole moment is

  1. T m
  2. N A⁻¹ m⁻¹
  3. A m⁻¹
  4. A m² Correct answer.

Explanation

For a coil of N turns, area A and current I, the magnetic moment is m = NIA, so its unit is ampere metre squared. The same unit can be written as joule per tesla, since the torque is mB sinθ.