Q.An electric dipole with dipole moment $4 \times 10^{-9}\,\text{C m}$ is aligned at $30^\circ$ with the direction of a uniform electric field of magnitude $5 \times 10^{4}\,\text{N C}^{-1}$. Calculate the magnitude of the torque acting on the dipole.

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A Magnetic needle lying parallel to a magnetic field requires W units of work to turn it through $60^\circ$. The torque required to keep the needle in this position will be (A) $\sqrt{3}W$ (B) W (C) $\dfrac{\sqrt{3}W}{2}$ (D) 2W

   ›Reveal solutionSolution

   This tests the work-torque relation for a magnetic dipole in a field; solving for MB from the given work and then finding torque at 60° gives $\sqrt{3}W$.

   Concept and Intuition

   A magnetic dipole in a field has potential energy $U=-MB\cos\theta$, and the work needed to rotate it from one orientation to another equals the change in this potential energy. Once we know $MB$ from the given work-to-rotate value, we can find the torque at the final orientation directly from $\tau = MB\sin\theta$ — the two formulas share the same $MB$ product, so the given work data effectively calibrates the torque formula for us.

   Step-by-Step Solution

   1. Work to rotate a dipole from angle $\theta_1$ to $\theta_2$: $W = MB(\cos\theta_1 - \cos\theta_2)$.
   2. Here $\theta_1 = 0$ (parallel to field) and $\theta_2 = 60^\circ$: $W = MB(\cos0^\circ - \cos60^\circ) = MB(1 - 0.5) = 0.5\,MB$.
   3. So $MB = 2W$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Two identical bar magnets are placed one above the other such that they are mutually perpendicular and bisect each other. The time period of this combination in a horizontal magnetic field is 'T'. The time period of each magnet in the same field is (A) $\sqrt{2}T$ (B) $2^{(1/4)}T$ (C) $2^{-(1/3)}T$ (D) $2^{-(1/4)}T$

   ›Reveal solutionSolution

   Crossing two identical bar magnets combines their moments as a perpendicular vector sum and their moments of inertia additively; working through the vibration-magnetometer period formula gives $T_{single}=2^{-1/4}T$. Answer: (D).

   Concept and Intuition

   In a vibration magnetometer, the period is $T=2\pi\sqrt{\dfrac{I}{MB}}$, where $I$ is the moment of inertia about the (vertical) oscillation axis and $M$ is the magnetic moment. When two identical magnets are crossed at right angles through their common centre, their magnetic moments (equal magnitude $M$, perpendicular directions) add as vectors to give $M'=\sqrt{M^2+M^2}=\sqrt2\,M$. Their moments of inertia about the vertical axis simply add, $I'=2I$, because for a thin rod lying in a horizontal plane, the moment of inertia about a vertical axis through its centre is $mL^2/12$ regardless of which horizontal direction the rod points — so orientation doesn't matter, only that both rods contribute independently.

   Step-by-Step Solution

   1. Single magnet period: $T_{single}=2\pi\sqrt{\dfrac{I}{MB}}$.
   2. Combination: $I'=2I$, $M'=\sqrt2\,M$, so $T_{combo}=2\pi\sqrt{\dfrac{2I}{\sqrt2\,MB}}=2\pi\sqrt{\dfrac{I}{MB}}\cdot\sqrt{\dfrac{2}{\sqrt2}}$. …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A magnetic needle kept in a non-uniform magnetic field experiences (A) torque but not force (B) neither torque nor force (C) both force and torque (D) force but not torque

   ›Reveal solutionSolution

   Tests the distinction between a magnetic dipole's behaviour in uniform vs non-uniform fields — torque always tends to align it, but a net translational force only appears when the field varies in space.

   Concept and Intuition

   A magnetic dipole (like a compass needle) has two "poles" experiencing forces from the external field. In a uniform field, these two forces are equal and opposite (same field strength at both poles), so they cancel to give zero net force — but they act at different points, producing a torque that twists the needle to align with the field. In a non-uniform field, the field strength differs at the two poles, so the two forces no longer cancel, leaving a net force in addition to the torque.

   Step-by-Step Solution

   1. Model the needle as a magnetic dipole with two poles of strength $\pm q_m$ separated by a small distance (dipole length $2l$), placed in field $\vec B(\vec r)$.
   2. Torque on a dipole: $\vec\tau = \vec m\times\vec B$, which is nonzero whenever the dipole moment $\vec m$ is not aligned with $\vec B$ — this exists in both uniform and non-uniform fields.
   3. Net force on a dipole: $\vec F = (\vec m\cdot\nabla)\vec B$ (equivalently, force is nonzero only where $\vec B$ varies with position, i.e. $\nabla B\neq0$).
   4. In a uniform field, $\nabla B = 0$, so $\vec F = 0$ — only torque acts. …

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4. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A rectangular coil of sides 6 cm & 5 cm respectively has 100 turns in it. It carries a current of 5A and is placed in a uniform magnetic field of 0.4 T, in such a manner that its plane makes an angle of $60°$ with the field direction. The torque on the coil is (A) $5 \times 10^{-2}\ N.m$ (B) 0.06 $N.m$ (C) 0.3 $N.m$ (D) 0.1 $N.m$

   ›Reveal solutionSolution

   Torque on a current loop is $\tau=NIAB\sin\theta$ where $\theta$ is the angle between the magnetic moment (normal to the coil) and $B$; here that angle is $30°$, giving $\tau=0.3$ N·m.

   Concept and Intuition

   The magnetic moment of a current loop points along its normal, not along its plane. So if the plane makes an angle of $60°$ with $\vec B$, the normal (and hence the moment vector) makes the complementary angle, $90°-60°=30°$, with $\vec B$ — and it's this angle that enters the torque formula $\tau=NIAB\sin\theta$.

   Step-by-Step Solution

   1. Area: $A = 0.06\,m \times 0.05\,m = 3\times10^{-3}\,m^2$.
   2. Angle between normal and field: $\theta = 90° - 60° = 30°$.
   3. Torque: $\tau = NIAB\sin\theta = 100 \times 5 \times 3\times10^{-3} \times 0.4 \times \sin30°$. …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A magnetic needle oscillating in a horizontal plane has a time period 2 s and 3 s at places where the angles of dip are $30°$ and $60°$ respectively. The ratio of magnetic fields at the two places is (A) $\dfrac{4\sqrt{3}}{7}$ (B) $\dfrac{4}{9\sqrt{3}}$ (C) $\dfrac{9}{4\sqrt{3}}$ (D) $\dfrac{9}{\sqrt{3}}$

   ›Reveal solutionSolution

   The oscillation period of a dip needle depends on the horizontal component of the field, $B_H=B\cos\delta$; combining the two given periods and dip angles gives $B_1/B_2 = 9/(4\sqrt3)$.

   Concept and Intuition

   A magnetic needle free to oscillate in a horizontal plane is restored only by the horizontal component of the Earth's field (the vertical component just tilts the axis, contributing no horizontal torque). So its period follows $T = 2\pi\sqrt{I/(mB_H)}$, i.e. $T\propto 1/\sqrt{B_H}$, and $B_H = B\cos\delta$ where $\delta$ is the angle of dip and $B$ is the total field at that place.

   Step-by-Step Solution

   1. $T \propto \dfrac{1}{\sqrt{B_H}} \Rightarrow \dfrac{T_1^2}{T_2^2} = \dfrac{B_{H2}}{B_{H1}}$.
   2. Given $T_1=2\,s$ (at $\delta_1=30°$), $T_2=3\,s$ (at $\delta_2=60°$): $\dfrac{4}{9} = \dfrac{B_{H2}}{B_{H1}} = \dfrac{B_2\cos60°}{B_1\cos30°}$. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.$1\mu C, -1\mu C$ charges are placed at a distance of 5 cm in forming a dipole. The amount of torque required to place this dipole perpendicular to an electric field of $3\times10^5\ \mathrm{NC^{-1}}$ is given by (A) $5\times10^{-3}\ N.m$ (B) $15\times10^{-3}\ N.m$ (C) $1\times10^{-3}\ N.m$ (D) $10\times10^{-3}\ N.m$

   ›Reveal solutionSolution

   The torque on a dipole is $\tau=pE\sin\theta$, maximized at $\theta=90^\circ$; computing $p$ from the given charges and separation gives $\tau=15\times10^{-3}\,N.m$.

   Concept and Intuition

   An electric dipole in a uniform field $E$ experiences a torque $\tau=\vec p\times\vec E$, with magnitude $pE\sin\theta$ where $\theta$ is the angle between the dipole moment and the field. This torque is zero when the dipole is aligned with the field and maximum ($\tau=pE$) when it is perpendicular to the field — which is exactly the orientation asked about here.

   Step-by-Step Solution

   1. Dipole moment: $p=q\times d=(1\times10^{-6}\,C)\times(0.05\,m)=5\times10^{-8}\,C\cdot m$.
   2. To hold the dipole perpendicular to the field ($\theta=90^\circ$), the required torque equals the maximum torque: $\tau=pE\sin90^\circ=pE$. …

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7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A short bar magnet is placed in a uniform magnetic field of 2 T such that the axis of the magnet makes an angle of $45^{\circ}$ with the direction of the magnetic field. If the torque acting on the magnet is $0.36\sqrt{2}\ \mathrm{N\,m}$, then the moment of the magnet is (A) $0.54\ \mathrm{J\,T^{-1}}$ (B) $0.18\ \mathrm{J\,T^{-1}}$ (C) $0.72\ \mathrm{J\,T^{-1}}$ (D) $0.36\ \mathrm{J\,T^{-1}}$

   ›Reveal solutionSolution

   Using the magnetic torque formula $\tau = MB\sin\theta$ and solving for $M$ gives $0.36\ \text{J T}^{-1}$.

   Concept and Intuition

   A bar magnet of magnetic moment $M$ placed in a uniform field $B$ experiences a torque that tends to align its axis with the field: $\tau = MB\sin\theta$, where $\theta$ is the angle between the magnetic axis and the field direction. This is directly analogous to the torque on an electric dipole in a uniform electric field.

   Step-by-Step Solution

   1. Given: $B = 2\ \text{T}$, $\theta = 45^\circ$, $\tau = 0.36\sqrt{2}\ \text{N m}$.
   2. $\sin 45^\circ = \dfrac{1}{\sqrt2} = \dfrac{\sqrt2}{2}$.
   3. From $\tau = MB\sin\theta$: $M = \dfrac{\tau}{B\sin\theta} = \dfrac{0.36\sqrt2}{2 \times \frac{\sqrt2}{2}} = \dfrac{0.36\sqrt2}{\sqrt2}$. …

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8. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.A wire of length 10 m carrying current of 1 A is bent into a circular loop. If a magnetic field of $2\pi \times 10^{-4}$ T is applied on the loop, then the maximum torque acting on it is (A) $100 \times 10^{-4}$ N m (B) $50 \times 10^{-4}$ N m (C) $25 \times 10^{-4}$ N m (D) $75 \times 10^{-4}$ N m

   ›Reveal solutionSolution

   Bending the given wire into a single circular loop fixes its radius (and hence area) from the total length; the maximum torque $\tau_{max}=NIAB$ then works out to $50\times10^{-4}$ N m.

   Concept and Intuition

   A current loop in a magnetic field experiences a torque $\tau = NIAB\sin\phi$, where $\phi$ is the angle between the loop's magnetic-moment vector (normal to the plane) and $B$. This torque is maximum when the plane of the loop is parallel to $B$ (i.e. $\phi=90°$, so $\sin\phi=1$), giving $\tau_{max}=NIAB$. Here, the wire's total length is fixed at 10 m and it is bent into one single loop (a natural reading unless stated otherwise), so that length becomes the loop's circumference, which pins down its radius and area — larger radius means larger area and hence larger torque for the same current and field.

   Step-by-Step Solution

   1. The wire of length $L=10$ m forms one circular loop, so $2\pi r = 10 \Rightarrow r = \dfrac{5}{\pi}$ m.
   2. Area of the loop: $A = \pi r^2 = \pi\left(\dfrac{5}{\pi}\right)^2 = \dfrac{25}{\pi}\ \mathrm{m^2}$.
   3. Magnetic moment: $m = NIA = 1\times1\times\dfrac{25}{\pi} = \dfrac{25}{\pi}\ \mathrm{A\,m^2}$. …

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9. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.An electric dipole with dipole moment $2\times10^{-10}\ \mathrm{C\,m}$ is aligned at an angle $30^{\circ}$ with the direction of uniform electric field of $10^4\ \mathrm{N\,C^{-1}}$. The magnitude of the torque acting on the dipole is (A) $10^{-6}\ \mathrm{N\,m}$ (B) $10^{-5}\ \mathrm{N\,m}$ (C) $10^{-4}\ \mathrm{N\,m}$ (D) $10^{-3}\ \mathrm{N\,m}$

   ›Reveal solutionSolution

   This tests the torque formula for a dipole in a uniform electric field, $\tau = pE\sin\theta$; direct substitution gives $10^{-6}$ N m.

   Concept and Intuition

   A dipole placed at an angle to a uniform field experiences a torque that tries to align it with the field — the two equal and opposite forces on the charges form a couple. The turning effect is maximum when the dipole is perpendicular to the field ($\theta = 90^\circ$) and zero when aligned with it, which is captured by $\tau = pE\sin\theta$.

   Step-by-Step Solution

   1. Given: $p = 2\times10^{-10}\,\mathrm{C\,m}$, $E = 10^4\,\mathrm{N\,C^{-1}}$, $\theta = 30^\circ$.
   2. $\tau = pE\sin\theta = (2\times10^{-10})(10^4)(\sin30^\circ)$. …

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10. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
    Q.A bar magnet of moment $0.4\times10^{-3}\ \mathrm{Am}^2$ is kept in a magnetic field of $2\pi\times10^{-3}$ T. The magnet makes an angle of $45°$ with the direction of magnetic field. The torque acting on the magnet is (A) $7.65\pi\times10^{-7}$ Nm (B) $6.55\pi\times10^{-4}$ Nm (C) $5.65\pi\times10^{-2}$ Nm (D) $5.65\pi\times10^{-7}$ Nm

    ›Reveal solutionSolution

    This tests the torque formula on a magnetic dipole, $\tau=mB\sin\theta$, with careful handling of the powers of ten and the $\sin45^\circ=\sqrt2/2$ factor.

    Concept and Intuition

    A bar magnet placed in a magnetic field at an angle experiences a torque that tends to align it with the field, given by $\tau=mB\sin\theta$ — directly analogous to the torque on an electric dipole in an electric field. The torque is maximum when the magnet is perpendicular to the field ($\theta=90^\circ$) and zero when aligned with it; at $45^\circ$ it's at an intermediate fraction ($\sin45^\circ=\sqrt2/2\approx0.707$) of that maximum.

    Step-by-Step Solution

    1. Torque formula: $\tau=mB\sin\theta$, with $m=0.4\times10^{-3}\ \text{Am}^2$, $B=2\pi\times10^{-3}\ \text{T}$, $\theta=45^\circ$.
    2. Compute $mB=(0.4\times10^{-3})(2\pi\times10^{-3})=0.8\pi\times10^{-6}$. …

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11. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.A current carrying coil experiences a torque due to a magnetic field. The value of the torque is 80 % of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is (A) $30°$ (B) $45°$ (C) $\tan^{-1}\left(\dfrac{3}{4}\right)$ (D) $\tan^{-1}\left(\dfrac{4}{3}\right)$

    ›Reveal solutionSolution

    The torque on a current loop is $\tau=\tau_{max}\sin\theta$ with $\theta$ measured from the coil's normal; setting $\sin\theta=0.8$ gives $\theta=\tan^{-1}(4/3)$.

    Concept and Intuition

    A current loop's magnetic moment $\vec m$ points along its normal. The torque $\vec\tau=\vec m\times\vec B$ has magnitude $mB\sin\theta$, where $\theta$ is the angle between $\vec m$ (the normal) and $\vec B$ — exactly as phrased in the question.

    Step-by-Step Solution

    1. $\tau=\tau_{max}\sin\theta$, given $\tau=0.8\,\tau_{max}$.
    2. $\sin\theta = 0.8 = \dfrac{4}{5}$, forming a 3-4-5 right triangle: $\cos\theta=\dfrac35$, $\tan\theta=\dfrac43$.
    3. $\theta = \tan^{-1}\!\left(\dfrac{4}{3}\right)$.

    Common Mistakes …

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12. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.A magnet suspended in a uniform magnetic field is heated so as to reduce its magnetic moment by 19%. By doing this, the time period of the magnet approximately (A) Increases by 11% (B) Decreases by 19% (C) Increases by 19% (D) Decreases by 4%

    ›Reveal solutionSolution

    Time period of an oscillating magnet is inversely proportional to the square root of its magnetic moment; a 19% drop in moment raises the period by about 11%.

    Concept and Intuition

    A freely-oscillating bar magnet in a uniform field behaves like a torsional pendulum, with $T=2\pi\sqrt{I/(MB)}$. Reducing $M$ reduces the "restoring strength," so the magnet oscillates more slowly, i.e. $T$ increases.

    Step-by-Step Solution

    1. $T\propto \dfrac{1}{\sqrt M}$ (with $I$, $B$ unchanged).
    2. New moment: $M' = M(1-0.19)=0.81M$. …

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