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 29 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If an electron moving with a velocity of $4 \times 10^6~\mathrm{ms}^{-1}$ enters a uniform magnetic field of $\dfrac{\pi}{2}~\mathrm{mT}$ at an angle of $60^\circ$ with the direction of the magnetic field, then the pitch of the helical path of the electron is (Mass of the electron $= 9 \times 10^{-31}~\mathrm{kg}$) (A) $1.5~\mathrm{cm}$ (B) $3~\mathrm{cm}$ (C) $4.5~\mathrm{cm}$ (D) $6~\mathrm{cm}$

   ›Reveal solutionSolution

   The pitch of a helix in a magnetic field depends only on the velocity component parallel to the field and the time period of circular motion. Here it comes out to 4.5 cm.

   When a charged particle enters a magnetic field at an angle that is not 0° or 90°, its velocity splits into two independent components: one parallel to the field ($v_{\parallel}$) and one perpendicular to it ($v_{\perp}$). The perpendicular component causes circular motion (Lorentz force provides centripetal force), while the parallel component is unaffected by the field and carries the particle forward uniformly. The combination gives a helical path.

   The pitch is the distance the particle moves along the field direction in one complete circular revolution. So pitch = (parallel speed) × (time period of one revolution).

   1. Resolve the velocity.

      The electron enters at $60^\circ$ to the magnetic field.

      $v = 4 \times 10^6~\mathrm{m/s}$

      $v_{\parallel} = v \cos 60^\circ = 4 \times 10^6 \times \frac{1}{2} = 2 \times 10^6~\mathrm{m/s}$

      $v_{\perp} = v \sin 60^\circ = 4 \times 10^6 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \times 10^6~\mathrm{m/s}$

   2. Find the time period of the circular motion.

      For a charge $q$ in a uniform magnetic field $B$, the time period is independent of speed:

      $T = \frac{2\pi m}{qB}$

      Here $B = \frac{\pi}{2}~\mathrm{mT} = \frac{\pi}{2} \times 10^{-3}~\mathrm{T}$, $m = 9 \times 10^{-31}~\mathrm{kg}$, $q = 1.6 \times 10^{-19}~\mathrm{C}$.

      $T = \frac{2\pi \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times \frac{\pi}{2} \times 10^{-3}}$

      Cancel $\pi$: …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If a thin conducting wire of length 12 m carrying a current of $2\sqrt{3}$ A is bent into a regular hexagonal loop and is placed in a uniform magnetic field of 2T, then the maximum torque acting on the loop is (A) 36 N m (B) 48 N m (C) 72 N m (D) 24 N m

   ›Reveal solutionSolution

   The maximum torque on a current loop is $\tau_{\max} = N I A B$, where $A$ is the area of the loop. For a regular hexagon of side $2$ m, area is $6\sqrt{3}$ m², giving $\tau_{\max} = 2\sqrt{3} \cdot 6\sqrt{3} \cdot 2 = 72$ N·m. The correct option is (C).

   Concept & Intuition

   When a current-carrying loop is placed in a uniform magnetic field, it experiences a torque that tends to rotate it so that its plane becomes perpendicular to the field. The magnitude of torque is $\tau = N I A B \sin\theta$, where $\theta$ is the angle between the magnetic field and the normal to the loop’s plane. The maximum torque occurs when $\sin\theta = 1$, i.e., when the field is parallel to the plane of the loop. So the problem reduces to finding the area of the hexagon formed by the wire.

   Step-by-step solution

   1. Determine the side length of the hexagon The wire is 12 m long and is bent into a regular hexagon. A regular hexagon has 6 equal sides.

   $\text{Side length } s = \frac{12}{6} = 2 \text{ m}.$

   2. Find the area of a regular hexagon A regular hexagon can be divided into 6 equilateral triangles, each of side $s$. The area of one equilateral triangle is $\frac{\sqrt{3}}{4} s^2$.

   $\text{Area of hexagon } A = 6 \times \frac{\sqrt{3}}{4} s^2 = \frac{6\sqrt{3}}{4} \times (2)^2 = \frac{6\sqrt{3}}{4} \times 4 = 6\sqrt{3} \text{ m}^2.$

   3. Apply the torque formula …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.A magnetic dipole is suspended in a region where two uniform magnetic fields are inclined at $75^\circ$ with each other. In stable equilibrium, if the dipole makes an angle of $30^\circ$ with the direction of one field of $10\sqrt{2}$ mT, then the value of the second magnetic field is (A) $20\sqrt{2}$ mT (B) $10\sqrt{2}$ mT (C) $10$ mT (D) $20$ mT

   ›Reveal solutionSolution

   At equilibrium the torques from the two fields balance: $B_1\sin 30^\circ = B_2\sin 45^\circ$, giving $B_2 = 10\,\mathrm{mT}$.

   Setup. The dipole sits in two uniform fields inclined at $75^\circ$. It makes $30^\circ$ with the first field $B_1 = 10\sqrt{2}\,\mathrm{mT}$, hence

   $75^\circ - 30^\circ = 45^\circ$

   with the second field $B_2$.

   Equilibrium condition. The torque exerted by each field on the dipole $m$ must cancel:

   $m B_1 \sin\theta_1 = m B_2 \sin\theta_2,$

   $B_1 \sin 30^\circ = B_2 \sin 45^\circ.$ …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the magnetic field at the equator of the earth is 0.4 G, then the earth's magnetic dipole moment (in $\mathrm{A\,m^2}$) is of the order of (Radius of the Earth = 6400 km) (A) $10^{18}$ (B) $10^{17}$ (C) $10^{26}$ (D) $10^{23}$

   ›Reveal solutionSolution

   The Earth’s magnetic field at the equator is modeled as that of a dipole at the centre. Using the formula for the equatorial field of a dipole, we solve for the dipole moment and find it is of order $10^{23} \, \mathrm{A\,m^2}$, so the correct option is (D).

   Concept & Intuition

   The Earth’s magnetic field, to a good approximation, behaves like a giant bar magnet at its centre — a magnetic dipole. At the equator, the field lines are horizontal and point north. For a dipole, the field strength at a point on the equator (perpendicular to the dipole axis) is half the strength at the pole at the same distance. The formula is simple:

   $B_{\text{eq}} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$

   where $M$ is the dipole moment, $r$ is the distance from the centre (Earth’s radius), and $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{T\,m/A}$. Given $B_{\text{eq}}$ and $r$, we can solve for $M$.

   Step-by-step solution

   1. Write the known quantities in SI units
      • Magnetic field at equator: $B = 0.4 \, \text{G}$ Recall $1 \, \text{G} = 10^{-4} \, \text{T}$, so

   $B = 0.4 \times 10^{-4} = 4 \times 10^{-5} \, \text{T}.$

   • Earth’s radius: $R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m}$.

   2. Recall the dipole field formula for the equatorial point For a magnetic dipole of moment $M$, the field at a distance $r$ on the equatorial plane is

   $B = \frac{\mu_0}{4\pi} \frac{M}{r^3}.$

   (This is derived from the general dipole field: $B = \frac{\mu_0}{4\pi} \frac{M}{r^3} \sqrt{1+3\cos^2\theta}$, where $\theta = 90^\circ$ at the equator gives $\cos\theta = 0$, so the factor becomes 1.)

   3. Solve for the dipole moment $M$ Rearranging:

   $M = \frac{4\pi}{\mu_0} \, B \, r^3.$

   Substitute $\mu_0 = 4\pi \times 10^{-7}$:

   $\frac{4\pi}{\mu_0} = \frac{4\pi}{4\pi \times 10^{-7}} = 10^7.$

   So

   $M = 10^7 \times B \times r^3.$

   4. Plug in numbers

   $B = 4 \times 10^{-5} \, \text{T}, \quad r = 6.4 \times 10^6 \, \text{m}.$

   First compute $r^3$:

   $r^3 = (6.4 \times 10^6)^3 = 6.4^3 \times 10^{18} = 262.144 \times 10^{18} \approx 2.62 \times 10^{20}.$

   Then …

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5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The average force exerted on $5\,\mathrm{cm}^2$ area of a non-reflecting plate in 10 minutes when light with an energy flux of $20\,\mathrm{W\,cm}^{-2}$ incidents normally on it is (A) $3.33 \times 10^{-7}\,\mathrm{N}$ (B) $6.66 \times 10^{-4}\,\mathrm{N}$ (C) $2.5 \times 10^{-7}\,\mathrm{N}$ (D) $5.55 \times 10^{-10}\,\mathrm{N}$

   ›Reveal solutionSolution

   For a non-reflecting (perfectly absorbing) surface, the radiation pressure equals the energy flux divided by the speed of light. The force is pressure times area, independent of time. The answer is $3.33 \times 10^{-7}\,\mathrm{N}$, option (A).

   The key concept here is radiation pressure — the force exerted by electromagnetic waves on a surface. When light hits a surface, it carries momentum. If the surface absorbs the light (non-reflecting), all that momentum is transferred to the surface, giving a force. If it reflected, the momentum change would be double, but here it's absorption.

   The energy flux (intensity) $I = 20\,\mathrm{W\,cm}^{-2}$ tells us how much energy arrives per second per square centimetre. Since light's momentum is related to its energy by $p = E/c$, the momentum delivered per second per unit area is $I/c$. That's exactly the radiation pressure for a perfectly absorbing surface.

   Now let's work it through step by step.

   1. Convert area to SI units. The area is $5\,\mathrm{cm}^2$. Since $1\,\mathrm{cm}^2 = 10^{-4}\,\mathrm{m}^2$, we have

   $A = 5 \times 10^{-4}\,\mathrm{m}^2.$

   2. Convert energy flux to SI units. The flux is $20\,\mathrm{W\,cm}^{-2}$. Since $1\,\mathrm{W\,cm}^{-2} = 10^4\,\mathrm{W\,m}^{-2}$,

   $I = 20 \times 10^4 = 2 \times 10^5\,\mathrm{W\,m}^{-2}.$

   3. Find the radiation pressure. For a perfectly absorbing surface,

   $P = \frac{I}{c},$

   where $c = 3 \times 10^8\,\mathrm{m\,s}^{-1}$. So

   $P = \frac{2 \times 10^5}{3 \times 10^8} = \frac{2}{3} \times 10^{-3} = 6.67 \times 10^{-4}\,\mathrm{Pa}.$

   $P_{\text{absorb}} = \frac{I}{c}$ …

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6. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.An electric dipole of length 4 cm is placed in a uniform electric field of intensity $2 \times 10^5~\mathrm{NC}^{-1}$. When the dipole is oriented at certain angle with the electric field, it experiences a torque of $0.2\sqrt{3}~\mathrm{Nm}$ and possesses a potential energy of $-0.2~\mathrm{J}$. The magnitude of each charge of the dipole is (A) $75~\mu\mathrm{C}$ (B) $50~\mu\mathrm{C}$ (C) $100~\mu\mathrm{C}$ (D) $25~\mu\mathrm{C}$

   ›Reveal solutionSolution

   The problem gives torque and potential energy for a dipole in a uniform field. Using the formulas $\tau = pE\sin\theta$ and $U = -pE\cos\theta$, we can solve for the dipole moment $p$ and then the charge $q = p/d$. The magnitude of each charge is $50~\mu\mathrm{C}$.

   The key idea is that torque and potential energy of a dipole in a uniform field are two sides of the same coin — both depend on the dipole moment $p$, the field $E$, and the orientation angle $\theta$. Given both quantities, we can eliminate $\theta$ and find $p$ directly.

   1. Write the two equations.

      For a dipole with moment $p$ in a uniform field $E$, the torque is $\tau = pE\sin\theta$ and the potential energy is $U = -pE\cos\theta$. Here $\theta$ is the angle between the dipole axis and the field direction.

      We are given:

      $\tau = 0.2\sqrt{3}~\mathrm{Nm}$,

      $U = -0.2~\mathrm{J}$,

      $E = 2 \times 10^5~\mathrm{NC}^{-1}$.

   2. Square and add to eliminate $\theta$.

      Notice that $(pE)^2 = (pE\sin\theta)^2 + (pE\cos\theta)^2 = \tau^2 + (-U)^2$, because $U = -pE\cos\theta$ so $pE\cos\theta = -U$.

      So:

   $(pE)^2 = \tau^2 + U^2$

   Substitute the numbers:

   $\tau^2 = (0.2\sqrt{3})^2 = 0.04 \times 3 = 0.12$

   $U^2 = (-0.2)^2 = 0.04$

   Hence:

   $(pE)^2 = 0.12 + 0.04 = 0.16$

   So $pE = \sqrt{0.16} = 0.4$ (taking the positive root, since $p$ and $E$ are magnitudes).

   3. Find the dipole moment $p$.

   $p = \frac{0.4}{E} = \frac{0.4}{2 \times 10^5} = 2 \times 10^{-6}~\mathrm{C\,m}$

   That is $p = 2~\mu\mathrm{C\,m}$.

   4. Relate $p$ to the charge $q$. …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.As shown in the figure, a uniform straight wire of length $30\sqrt{3}$ cm is bent in the form of an equilateral triangle ABC. A uniform magnetic field 2T is applied parallel to the side BC. If the current through the wire is 2A, the magnitude of the force on the side AC is ($\vec{B}$ represents the direction of the magnetic field) (A) $2\sqrt{3}$ N (B) $0.2\sqrt{3}$ N (C) $1.2$ N (D) $0.6$ N

   ›Reveal solutionSolution

   The force on side AC is found by applying the magnetic force formula $F = I L B \sin\theta$ to the segment AC, using the geometry of the equilateral triangle and the given field direction. The result is $0.6$ N, so the correct option is (D).

   Concept & Intuition

   The magnetic force on a current-carrying wire in a uniform field is given by $\vec{F} = I \, \vec{L} \times \vec{B}$, where $\vec{L}$ is a vector along the wire in the direction of the current. The magnitude is $F = I L B \sin\theta$, with $\theta$ the angle between the wire and the magnetic field. Here, the field is parallel to side BC, so we must find the angle between side AC and the field direction. The wire is bent into an equilateral triangle, so all sides are equal and all internal angles are $60^\circ$. The key is to correctly identify the angle between AC and the field (which is along BC).

   Step-by-step solution

   1. Determine the side length of the triangle The total length of the wire is $30\sqrt{3}$ cm. An equilateral triangle has three equal sides, so each side length is

   $L = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ cm} = 0.1\sqrt{3} \text{ m}.$

   (Convert to meters for SI units: $1$ cm = $0.01$ m.)

   2. Identify the direction of the magnetic field

      The field $\vec{B}$ is parallel to side BC. In an equilateral triangle, side BC is one of the sides. So the field direction is along BC.

   3. Find the angle between side AC and the field

      In an equilateral triangle, each interior angle is $60^\circ$. Side AC meets side BC at vertex C. The angle between side AC and side BC is exactly the interior angle at C, which is $60^\circ$.

      Therefore, the angle $\theta$ between the current direction along AC and the magnetic field (along BC) is $60^\circ$.

   4. Apply the magnetic force formula

      The magnitude of the force on side AC is

      $F = I \, L \, B \, \sin\theta,$ …

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.As shown in the figure, a uniform straight wire of length $30\sqrt{3}$ cm is bent in the form of an equilateral triangle ABC. A uniform magnetic field 2T is applied parallel to the side BC. If the current through the wire is 2A, the magnitude of the force on the side AC is ($\vec{B}$ represents the direction of the magnetic field) (A) $0.2\sqrt{3}$ N (B) $1.2$ N (C) $0.6$ N (D) $2\sqrt{3}$ N

   ›Reveal solutionSolution

   The force on side AC is found by applying the magnetic force formula $\vec{F}=I\vec{L}\times\vec{B}$ to the segment AC, using the geometry of the equilateral triangle and the given field direction parallel to BC. The magnitude is $0.6$ N, so the correct option is (C).

   Concept & Intuition

   The magnetic force on a current-carrying wire in a uniform field is given by $\vec{F}=I\vec{L}\times\vec{B}$, where $\vec{L}$ is a vector along the wire in the direction of the current. The magnitude is $F = I L B \sin\theta$, with $\theta$ the angle between the wire and the field. Here, the field is parallel to side BC, so the force on each side depends on its orientation relative to BC. For side AC, we need its length and the angle between AC and the field direction.

   Step-by-step solution

   1. Find the side length of the equilateral triangle. The total wire length is $30\sqrt{3}$ cm, and it forms an equilateral triangle, so each side has length

   $L = \frac{30\sqrt{3}}{3} = 10\sqrt{3}\ \text{cm} = 0.1\sqrt{3}\ \text{m}.$

   2. Set up the geometry.

      Let triangle ABC be equilateral. Side BC is horizontal (say). The magnetic field $\vec{B}$ is parallel to BC, so we take $\vec{B}$ along the horizontal direction. Side AC makes an angle of $60^\circ$ with BC (since each interior angle is $60^\circ$). Therefore, the angle between the current direction along AC and the magnetic field is $60^\circ$.

   3. Apply the magnetic force formula to side AC.

      Current $I = 2$ A, magnetic field $B = 2$ T, length of AC = $0.1\sqrt{3}$ m, and $\theta = 60^\circ$.

      $$$$

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9. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.The charge on a parallel plate capacitor is $200~\mu \mathrm{C}$ and its capacitance is $4~\mu \mathrm{F}$. If the distance between the plates of the capacitor is $2~\mathrm{mm}$, then the electric force between the plates of the capacitor is (A) $2.5~\mathrm{N}$ (B) $5~\mathrm{N}$ (C) $10~\mathrm{N}$ (D) $1.25~\mathrm{N}$

   ›Reveal solutionSolution

   The attractive force between the plates is $F=\dfrac{Q^2}{2C d}$. Substituting $Q=200\,\mu\text{C}$, $C=4\,\mu\text{F}$, $d=2$ mm gives $F=2.5$ N (option A).

   Concept. Each plate feels the field produced by the other plate only. The field of a single plate is $\dfrac{\sigma}{2\varepsilon_0}$, so the force on the plate carrying charge $Q$ is

   $F=Q\cdot\frac{\sigma}{2\varepsilon_0}=\frac{Q^2}{2\varepsilon_0 A}.$

   Using $C=\dfrac{\varepsilon_0 A}{d}\Rightarrow\varepsilon_0 A=Cd$, this becomes

   $F=\frac{Q^2}{2Cd}.$

   Given data.

   $Q=200\,\mu\text{C}=2\times10^{-4}\text{ C},\quad C=4\,\mu\text{F}=4\times10^{-6}\text{ F},\quad d=2\text{ mm}=2\times10^{-3}\text{ m}.$

   Substitute. …

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10. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.If the magnetization of a bar magnet of area of cross-section $0.75\,\text{cm}^2$ and magnetic moment $2.7\,\text{Am}^2$ is $4 \times 10^5\,\text{Am}^{-1}$, then the length of the magnet is (A) $15\,\text{cm}$ (B) $9\,\text{cm}$ (C) $6\,\text{cm}$ (D) $12\,\text{cm}$

    ›Reveal solutionSolution

    Magnetization $M$ is magnetic moment per unit volume. Using $M = \frac{m}{V}$ and $V = A \cdot l$, we find the length $l = 9\,\text{cm}$.

    The key idea here is that magnetization ($M$) is defined as the magnetic moment per unit volume of the material. For a bar magnet, the volume is simply the product of its cross-sectional area and its length. So if you know the magnetic moment, the area, and the magnetization, you can directly solve for the length.

    A common mistake is to confuse magnetization with magnetic moment per unit length (which is the pole strength). Always remember: magnetization is a volume density, not a linear one.

    Let’s work through it step by step.

    1. Write down the definition of magnetization. Magnetization $M$ is given by

    $M = \frac{m}{V}$

    where $m$ is the magnetic moment and $V$ is the volume of the magnet.

    2. Express volume in terms of area and length. For a bar magnet of uniform cross-section,

    $V = A \cdot l$

    where $A$ is the area of cross-section and $l$ is the length.

    3. Substitute into the magnetization formula.

    $M = \frac{m}{A \cdot l}$

    4. Rearrange to solve for length $l$.

    $l = \frac{m}{M \cdot A}$

    5. Plug in the given values with consistent units.

       • Magnetic moment $m = 2.7\,\text{Am}^2$
       • Magnetization $M = 4 \times 10^5\,\text{Am}^{-1}$
       • Area $A = 0.75\,\text{cm}^2 = 0.75 \times 10^{-4}\,\text{m}^2 = 7.5 \times 10^{-5}\,\text{m}^2$

       So …

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11. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
    Q.A straight wire of length 90 cm carrying a current of 3 A is bent in the form of an equilateral triangular loop and is placed in a uniform magnetic field of $8 \times 10^{-4}$ T such that the plane of the loop makes an angle of $30^\circ$ with the direction of the magnetic field. The torque acting on the triangular loop is (A) $81\sqrt{3} \times 10^{-6}$ Nm (B) $27 \times 10^{-6}$ Nm (C) $27\sqrt{3} \times 10^{-6}$ Nm (D) $81 \times 10^{-6}$ Nm

    ›Reveal solutionSolution

    The torque on a current loop in a magnetic field depends on its magnetic moment, the magnetic field strength, and the sine of the angle between the magnetic moment vector (normal to the loop's plane) and the magnetic field. After calculating the area of the equilateral triangle and determining the correct angle, the torque is found to be $\boxed{81 \times 10^{-6} \text{ Nm}}$.

    When a current-carrying loop is placed in a uniform magnetic field, it experiences a torque. This torque tends to align the loop's magnetic moment with the direction of the magnetic field. The underlying principle is that the magnetic field exerts forces on the current elements within the loop. While the net force on a closed loop in a uniform magnetic field is zero, these forces can produce a net torque.

    The magnetic moment ($\vec{M}$) of a current loop is a vector quantity defined as the product of the current ($I$) and the area vector ($\vec{A}$) of the loop. The direction of the area vector is perpendicular to the plane of the loop, determined by the right-hand rule (if current flows counter-clockwise, $\vec{A}$ points out of the page).

    The magnetic moment of a single loop is given by $\vec{M} = I\vec{A}$. Its magnitude is $M = IA$.

    The torque ($\vec{\tau}$) experienced by a magnetic dipole (like a current loop) in a uniform magnetic field ($\vec{B}$) is given by the cross product of the magnetic moment and the magnetic field:

    $\vec{\tau} = \vec{M} \times \vec{B}$

    The magnitude of this torque is:

    $\tau = MB \sin\theta$

    where $\theta$ is the angle between the magnetic moment vector $\vec{M}$ and the magnetic field vector $\vec{B}$.

    Let's apply these concepts to solve the problem.

    1. Calculate the side length of the equilateral triangular loop:

       The total length of the wire is $90 \text{ cm}$. When bent into an equilateral triangle, this length is distributed equally among its three sides.

       Length of wire, $L = 90 \text{ cm} = 0.9 \text{ m}$.

       Side length of the triangle, $a = \frac{L}{3} = \frac{0.9 \text{ m}}{3} = 0.3 \text{ m}$.

    2. Calculate the area of the equilateral triangular loop:

       The area of an equilateral triangle with side length $a$ is given by the formula:

       $A = \frac{\sqrt{3}}{4} a^2$

       Substituting the value of $a$:

       $A = \frac{\sqrt{3}}{4} (0.3 \text{ m})^2 = \frac{\sqrt{3}}{4} (0.09 \text{ m}^2) = 0.0225\sqrt{3} \text{ m}^2$.

    3. Calculate the magnetic moment of the loop:

       The current flowing through the loop is $I = 3 \text{ A}$.

       The magnetic moment $M = IA$.

       $M = (3 \text{ A}) \times (0.0225\sqrt{3} \text{ m}^2) = 0.0675\sqrt{3} \text{ A m}^2$.

    4. Determine the angle between the magnetic moment and the magnetic field:

       The problem states that the plane of the loop makes an angle of $30^\circ$ with the direction of the magnetic field.

       The magnetic moment vector $\vec{M}$ is always perpendicular to the plane of the loop.

       If the plane makes an angle $\alpha = 30^\circ$ with the magnetic field $\vec{B}$, then the angle $\theta$ between the magnetic moment vector $\vec{M}$ (which is normal to the plane) and the magnetic field $\vec{B}$ is:

       $\theta = 90^\circ - \alpha = 90^\circ - 30^\circ = 60^\circ$. …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The potential difference between the ends of a straight conductor of length 20 cm is 16 V. If the drift speed of the electrons is $2.4 \times 10^{-4}$ ms$^{-1}$, the electron mobility in m$^2$ V$^{-1}$ s$^{-1}$ is (A) $3.6 \times 10^{-6}$ (B) $2.4 \times 10^{-6}$ (C) $2 \times 10^{-6}$ (D) $3 \times 10^{-6}$

    ›Reveal solutionSolution

    The key idea is that mobility $\mu = v_d / E$, where $E = V/L$. Using $V=16\ \text{V}$, $L=0.20\ \text{m}$, and $v_d = 2.4\times10^{-4}\ \text{m/s}$, we find $\mu = 3.0\times10^{-6}\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$, which corresponds to option (D).

    Concept & Intuition

    Electron mobility measures how quickly an electron drifts through a conductor in response to an electric field. The relation is simple:

    $\mu = \frac{v_d}{E}$

    where $v_d$ is the drift speed and $E$ is the electric field. The electric field in a uniform conductor is just the potential difference divided by its length: $E = V/L$. So the problem reduces to plugging in the given numbers — but careful with units!

    Step-by-step solution

    1. Identify the given quantities

       • Length of conductor: $L = 20\ \text{cm} = 0.20\ \text{m}$
       • Potential difference: $V = 16\ \text{V}$
       • Drift speed: $v_d = 2.4 \times 10^{-4}\ \text{m/s}$

    2. Find the electric field

       The electric field inside a straight conductor (assuming uniform field) is

    $E = \frac{V}{L} = \frac{16}{0.20} = 80\ \text{V/m}$

    3. Apply the mobility formula Mobility is defined as drift speed per unit electric field:

    $\mu = \frac{v_d}{E} = \frac{2.4 \times 10^{-4}}{80}$

    4. Calculate …

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