Q.A square loop of side $10\ \text{cm}$ and resistance $0.5\ \Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.10\ \text{T}$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $0.70\ \text{s}$ at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

Showing the 12 most recent of 37 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If a conducting rod of length 100 cm rotates about one of its ends with a constant frequency of 14 revolutions per second in a plane perpendicular to a uniform magnetic field of 2 T, then the induced emf between the two ends of the rod is (A) 144 V (B) 88 V (C) 122 V (D) 230 V

   ›Reveal solutionSolution

   A rotating rod in a magnetic field acts like a moving conductor; the induced emf is given by $\frac{1}{2} B \omega L^2$, yielding 88 V.

   The key idea here is motional emf. When a conductor moves through a magnetic field, free charges inside experience a magnetic force, which pushes them to one end until an electric field builds up to balance it. The potential difference that results is the induced emf. For a rod rotating about one end, different parts of the rod move at different speeds — the tip moves fastest, the pivot not at all — so we cannot simply use $Blv$ with a single $v$. Instead, we integrate the contribution from each infinitesimal segment.

   Let’s work it through.

   1. Set up the geometry and parameters. The rod length is $L = 100\ \text{cm} = 1\ \text{m}$. It rotates with frequency $f = 14\ \text{Hz}$, so its angular speed is

   $\omega = 2\pi f = 2\pi \times 14 = 28\pi\ \text{rad/s}.$

   The magnetic field is $B = 2\ \text{T}$, uniform and perpendicular to the plane of rotation.

   2. Consider a small element of the rod. Take a tiny segment of length $dr$ at a distance $r$ from the pivot. Its linear speed is $v = \omega r$. In a field perpendicular to both the rod and its motion, the motional emf across this segment is

   $d\mathcal{E} = B\, v\, dr = B\, (\omega r)\, dr.$

   This is the same as saying the magnetic force per unit charge on that segment is $vB$, and the potential difference across the segment is that field times $dr$.

   3. Integrate over the whole rod. The segments are in series along the rod, so the total emf between the ends is the sum (integral) of these infinitesimal contributions:

   $\mathcal{E} = \int_0^L B \omega r \, dr = B\omega \int_0^L r\, dr = B\omega \cdot \frac{L^2}{2}.$

   So the formula is

   $\mathcal{E} = \frac{1}{2} B \omega L^2.$ …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the number density of free electrons in a copper wire is $8.5 \times 10^{22}~\text{cm}^{-3}$ and the relaxation time of free electrons in the wire is $2.25 \times 10^{-14}~\text{s}$, then the electrical conductivity of copper is (Mass of the electron $= 9 \times 10^{-31}~\text{kg}$ and charge of the electron $= 1.6 \times 10^{-19}~\text{C}$) (A) $6.80 \times 10^{7}~\text{Sm}^{-1}$ (B) $3.40 \times 10^{7}~\text{Sm}^{-1}$ (C) $5.44 \times 10^{7}~\text{Sm}^{-1}$ (D) $2.72 \times 10^{7}~\text{Sm}^{-1}$

   ›Reveal solutionSolution

   Use $\sigma = \dfrac{n e^{2}\tau}{m}$ with the given values $\Rightarrow \sigma = 5.44 \times 10^{7}~\text{S m}^{-1}$.

   Given

   • Number density $n = 8.5 \times 10^{22}~\text{cm}^{-3} = 8.5 \times 10^{28}~\text{m}^{-3}$
   • Relaxation time $\tau = 2.25 \times 10^{-14}~\text{s}$
   • Electron mass $m = 9 \times 10^{-31}~\text{kg}$, charge $e = 1.6 \times 10^{-19}~\text{C}$

   Formula

   The electrical conductivity in the Drude model is

   $\sigma = \frac{n e^{2}\tau}{m}$

   Substitute

   $\sigma = \frac{(8.5 \times 10^{28})(1.6 \times 10^{-19})^{2}(2.25 \times 10^{-14})}{9 \times 10^{-31}}$ …

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3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.The relation between the total magnetic flux ($\phi$ in weber) linked with a coil of resistance $48~\Omega$ and the time (t in second) is $\phi = 5 - 24t$. If there is no loss of heat to the surroundings and the heat capacity of the coil is $5~\mathrm{JK}^{-1}$, then the rise in temperature of the coil during the time interval $t = 3~\mathrm{s}$ to $t = 5~\mathrm{s}$ is (A) $9.6~\mathrm{K}$ (B) $2.4~\mathrm{K}$ (C) $4.8~\mathrm{K}$ (D) $3.6~\mathrm{K}$

   ›Reveal solutionSolution

   The induced emf is constant, so the current and the rate of Joule heating are constant. The total heat produced over the 2 s interval divided by the heat capacity gives the temperature rise: 4.8 K.

   The problem gives you the magnetic flux linked with a coil as a function of time: $\phi = 5 - 24t$ (in weber). The coil has a resistance $R = 48\ \Omega$ and a heat capacity $C = 5\ \mathrm{J\,K^{-1}}$. You are asked for the rise in temperature from $t = 3\ \mathrm{s}$ to $t = 5\ \mathrm{s}$, assuming no heat loss.

   The key idea is that a changing flux induces an emf in the coil, which drives a current. That current dissipates energy as heat in the resistance (Joule heating). All of that heat goes into raising the coil’s temperature, since there is no loss to the surroundings.

   ---

   1. Find the induced emf. Faraday’s law says the induced emf is the negative rate of change of flux:

   $\mathcal{E} = -\frac{d\phi}{dt}.$

   Here $\phi = 5 - 24t$, so

   $\frac{d\phi}{dt} = -24.$

   Therefore

   $\mathcal{E} = -(-24) = 24\ \mathrm{V}.$

   The emf is constant — that’s important: it doesn’t change with time.

   2. Find the current in the coil. By Ohm’s law,

   $I = \frac{\mathcal{E}}{R} = \frac{24}{48} = 0.5\ \mathrm{A}.$

   This current is also constant over the entire interval.

   3. Calculate the power dissipated as heat. The rate of Joule heating (power) is

   $P = I^2 R = (0.5)^2 \times 48 = 0.25 \times 48 = 12\ \mathrm{W}.$

   Since $I$ and $R$ are constant, $P$ is constant.

   4. Find the total heat produced in the given time interval. The interval runs from $t = 3\ \mathrm{s}$ to $t = 5\ \mathrm{s}$, so the duration is $\Delta t = 2\ \mathrm{s}$. The total heat energy generated is

   $Q = P \times \Delta t = 12 \times 2 = 24\ \mathrm{J}.$

   5. Relate heat to temperature rise. …

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.At a point P on the axis of a current carrying circular coil of radius $R$, the magnetic field is $B$. If the distance of point P from the centre of the coil is $R\sqrt{15}$, then the magnetic field at a point Q on the axis of the coil which is at a distance of $R\sqrt{3}$ from the centre of the coil is (A) $B\sqrt{5}$ (B) $5B$ (C) $B\sqrt{8}$ (D) $8B$

   ›Reveal solutionSolution

   The magnetic field on the axis of a circular coil varies with distance from the center. By applying the formula for the magnetic field at two different points on the axis, we find that the magnetic field at point Q is $\boxed{8B}$.

   The magnetic field produced by a current-carrying circular coil at a point on its axis is a fundamental concept in electromagnetism. The strength of this field depends on several factors: the current flowing through the coil, the number of turns in the coil, the radius of the coil, and crucially, the distance of the point from the center of the coil along its axis.

   For a given coil, the current, number of turns, and radius are fixed. Therefore, the magnetic field strength at any point on its axis will vary only with the distance of that point from the coil's center. This problem requires us to use the formula for the axial magnetic field to relate the field strengths at two different distances from the coil's center. We will set up an expression for the magnetic field at each given point and then determine their ratio.

   1. Recall the formula for the magnetic field on the axis of a circular coil.

      The magnetic field $B_{axis}$ at a point on the axis of a circular coil of radius $R$, carrying current $I$, with $N$ turns, at a distance $x$ from its center is given by:

      $B_{axis} = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$

      where $\mu_0$ is the permeability of free space.

      For a specific coil, $\mu_0$, $N$, $I$, and $R$ are constants. We can group these constants together for simplicity. Let $K = \frac{\mu_0 N I R^2}{2}$. Then the formula becomes $B_{axis} = \frac{K}{(R^2 + x^2)^{3/2}}$.

   2. Calculate the magnetic field at point P.

      Point P is at a distance $x_P = R\sqrt{15}$ from the center of the coil. The magnetic field at P is given as $B$.

      Substituting $x_P$ into the formula:

   $B = \frac{K}{(R^2 + (R\sqrt{15})^2)^{3/2}}$

   $B = \frac{K}{(R^2 + 15R^2)^{3/2}}$

   $B = \frac{K}{(16R^2)^{3/2}}$

   To simplify $(16R^2)^{3/2}$, we can write it as $((4R)^2)^{3/2} = (4R)^3 = 64R^3$.
   So, the magnetic field at point P is:
   

   $B = \frac{K}{64R^3} \quad (*)$

   3. Calculate the magnetic field at point Q. …

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5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.The relative permeability of the material of the core of a solenoid is 400 and the number of turns per metre of the solenoid is 900. If the windings of the solenoid are insulated from the core and a current of 1.5 A is passed through the solenoid, then its magnetic intensity is (A) 1350 Am$^{-1}$ (B) 600 Am$^{-1}$ (C) 1.66 \times 10^{-3}$Am$^{-1}$(D) 0.678 Am$^{-1}$

   ›Reveal solutionSolution

   Magnetic intensity ($H$) inside a solenoid depends only on the current flowing through its windings and the number of turns per unit length, not on the magnetic properties of the core material. The magnetic intensity is $\boxed{\text{1350 Am}^{-1}}$.

   Concept and Intuition

   When a current flows through a solenoid, it creates a magnetic field. We describe this magnetic influence using two related quantities: magnetic field strength ($B$) and magnetic intensity ($H$).

   • Magnetic Intensity ($H$): This quantity represents the magnetizing effect of the external current alone. It's a measure of how strongly the current tries to magnetize the material. Crucially, $H$ depends only on the current and the geometry of the current-carrying conductor (like the number of turns per unit length in a solenoid). It does not depend on the material placed inside the solenoid. Think of it as the "effort" put in by the current to create magnetism.

   • Magnetic Field Strength ($B$): This quantity represents the actual magnetic field produced inside the material. It's the total magnetic field experienced by a point within the material. $B$ depends on both the external current (which determines $H$) and the magnetic properties of the material itself. The material's response to the magnetizing effort ($H$) is described by its permeability ($\mu$). So, $B = \mu H$.

   The problem asks for the magnetic intensity ($H$). Since $H$ is determined solely by the current and the solenoid's construction, the relative permeability of the core material is irrelevant for its calculation. It would be needed if we were asked to find the magnetic field strength ($B$) inside the core.

   Step-by-step Solution

   1. Identify the given parameters:

      We are given the following information:

      • Relative permeability of the core material, $\mu_r = 400$.
      • Number of turns per metre of the solenoid, $n = 900 \text{ m}^{-1}$.
      • Current passed through the solenoid, $I = 1.5 \text{ A}$.

   2. Recall the formula for magnetic intensity in a solenoid: …

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6. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.For a coil rotating in a uniform magnetic field, the phase difference between the flux linked with the coil and the current induced in it is (A) $90^\circ$ (B) $180^\circ$ (C) $60^\circ$ (D) $360^\circ$

   ›Reveal solutionSolution

   The flux linked with a rotating coil varies sinusoidally, and the induced current is proportional to the negative rate of change of flux. This derivative introduces a phase shift of $90^\circ$ between flux and current, so the correct answer is (A).

   The key here is to connect two ideas: how magnetic flux changes when a coil rotates, and how Faraday’s law tells us the induced emf (and hence current) depends on the rate of change of that flux. Many students jump straight to the formula without thinking about what “phase difference” actually means physically — it’s the angular shift between two sinusoidal quantities.

   Let’s break it down.

   1. Flux as a function of angle. When a coil of area $A$ rotates with angular speed $\omega$ in a uniform magnetic field $B$, the angle between the field and the coil’s normal changes as $\theta = \omega t$ (assuming it starts perpendicular to the field). The magnetic flux through the coil is

   $\Phi = B A \cos(\omega t).$

   This is a cosine wave — maximum when the coil is perpendicular to the field ($t=0$), zero when it’s parallel.

   2. Induced emf from Faraday’s law. The induced emf is the negative rate of change of flux:

   $\mathcal{E} = -\frac{d\Phi}{dt} = -\frac{d}{dt}[B A \cos(\omega t)] = B A \omega \sin(\omega t).$

   Notice: the derivative of $\cos$ gives $-\sin$, and the extra negative sign from Faraday’s law turns $-\sin$ into $+\sin$. So $\mathcal{E} \propto \sin(\omega t)$.

   3. Relating current to flux.

      For a purely resistive coil (which is the standard assumption here), the induced current $I$ is in phase with the emf: $I \propto \mathcal{E} \propto \sin(\omega t)$. So we compare $\Phi \propto \cos(\omega t)$ with $I \propto \sin(\omega t)$.

   4. Finding the phase difference. …

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Two identical wires, carrying equal currents are bent into circular coils A and B with 2 and 3 turns respectively. The ratio of the magnetic fields at the centres of the coils A and B is (A) $4 : 9$ (B) $2 : 3$ (C) $9 : 4$ (D) $3 : 2$

   ›Reveal solutionSolution

   The magnetic field at the centre of a circular coil depends on the number of turns and the radius. Since the wire length is fixed, more turns means a smaller radius, and the field scales as turns divided by radius. The ratio comes out to $9:4$, so option (C) is correct.

   Concept & Intuition

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

   $B = \frac{\mu_0 N I}{2r}.$

   Here, the two coils are made from identical wires of the same length. That means the total length of wire used for each coil is the same. If one coil has more turns, each turn must have a smaller circumference, hence a smaller radius. So the number of turns and the radius are linked: more turns → smaller radius. The ratio of fields therefore depends on how $N$ and $r$ trade off.

   ---

   Step-by-step solution

   1. Let the total length of wire be $L$ (same for both coils). For coil A with $N_A = 2$ turns, each turn has circumference $2\pi r_A$. So

   $L = N_A \cdot 2\pi r_A = 2 \cdot 2\pi r_A = 4\pi r_A.$

   For coil B with $N_B = 3$ turns,

   $L = N_B \cdot 2\pi r_B = 3 \cdot 2\pi r_B = 6\pi r_B.$

   2. Equate the lengths (since the wires are identical):

   $4\pi r_A = 6\pi r_B \quad\Rightarrow\quad r_A = \frac{3}{2} r_B.$

   So coil A has a larger radius than coil B.

   3. Write the magnetic field expressions:

   $B_A = \frac{\mu_0 N_A I}{2 r_A} = \frac{\mu_0 \cdot 2 \cdot I}{2 r_A} = \frac{\mu_0 I}{r_A},$

   $B_B = \frac{\mu_0 N_B I}{2 r_B} = \frac{\mu_0 \cdot 3 \cdot I}{2 r_B}.$

   4. Find the ratio $B_A : B_B$:

   $\frac{B_A}{B_B} = \frac{\mu_0 I / r_A}{\frac{3\mu_0 I}{2 r_B}} = \frac{1}{r_A} \cdot \frac{2 r_B}{3} = \frac{2 r_B}{3 r_A}.$

   Substitute $r_A = \frac{3}{2} r_B$:

   $\frac{B_A}{B_B} = \frac{2 r_B}{3 \cdot \frac{3}{2} r_B} = \frac{2 r_B}{\frac{9}{2} r_B} = \frac{2}{1} \cdot \frac{2}{9} = \frac{4}{9}.$ …

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8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $B_V$ and $B_H$ are respectively the vertical and horizontal components of the earth's magnetic field at a place where the angle of dip is $60^\circ$, then the total magnetic field at that place is (A) $\sqrt{3} B_H$ (B) $\sqrt{3} B_V$ (C) $\dfrac{2}{\sqrt{3}} B_V$ (D) $\dfrac{\sqrt{3}}{2} B_H$

   ›Reveal solutionSolution

   The total magnetic field is found from the dip angle using the relation $\tan \delta = B_V / B_H$. For $\delta = 60^\circ$, $B_V = \sqrt{3} B_H$, so the total field $B = \sqrt{B_H^2 + B_V^2} = 2 B_H = \frac{2}{\sqrt{3}} B_V$. The correct option is (C).

   The key idea is that Earth’s total magnetic field $\mathbf{B}$ at a place can be resolved into a horizontal component $B_H$ and a vertical component $B_V$. The angle of dip (or inclination) $\delta$ is the angle that the total field makes with the horizontal. This gives a right‑triangle relationship between $B$, $B_H$, and $B_V$.

   Why this works:

   If you picture the total field as the hypotenuse of a right triangle whose legs are $B_H$ (adjacent to $\delta$) and $B_V$ (opposite to $\delta$), then:

   • $\tan \delta = \frac{B_V}{B_H}$
   • $\cos \delta = \frac{B_H}{B}$
   • $\sin \delta = \frac{B_V}{B}$

   Given $\delta = 60^\circ$, we can use these to express $B$ in terms of either $B_H$ or $B_V$.

   1. Use the dip angle to relate $B_V$ and $B_H$.

      $\tan 60^\circ = \sqrt{3} = \frac{B_V}{B_H}$

      So $B_V = \sqrt{3} \, B_H$ and also $B_H = \frac{B_V}{\sqrt{3}}$.

   2. Find the total field $B$ using Pythagoras.

      $B = \sqrt{B_H^2 + B_V^2}$

      Substitute $B_V = \sqrt{3} B_H$:

      $B = \sqrt{B_H^2 + 3 B_H^2} = \sqrt{4 B_H^2} = 2 B_H$

      This matches none of the options directly, but we can rewrite it.

   3. Express $B$ in terms of $B_V$.

      From $B_H = \frac{B_V}{\sqrt{3}}$,

      $B = 2 \cdot \frac{B_V}{\sqrt{3}} = \frac{2}{\sqrt{3}} B_V$

      This is option (C). …

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9. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The potential difference between the terminals of a cell is 20 V when a current of 2 A flows through the circuit. When the direction of current in the circuit is reversed, the potential difference between the terminals of the cell is 30 V. The internal resistance of the cell is (A) $1~\Omega$ (B) $1.5~\Omega$ (C) $2~\Omega$ (D) $2.5~\Omega$

   ›Reveal solutionSolution

   The key idea is that the terminal voltage changes because the internal resistance drops voltage differently depending on whether the cell is discharging or being charged. Solving the two equations gives internal resistance $2.5\ \Omega$, so the correct option is (D).

   When a real cell supplies current, its terminal voltage is less than its emf because of the voltage drop across its internal resistance. But if you reverse the current — forcing current into the cell (charging it) — the internal resistance now adds to the terminal voltage. That difference in terminal voltages directly reveals the internal resistance.

   Let’s work it out step by step.

   1. Define the variables Let the cell’s emf be $\mathcal{E}$ and its internal resistance be $r$. When the cell delivers current $I$ (discharging), the terminal voltage $V$ is:

   $V = \mathcal{E} - I r$

   Here $I = 2\ \text{A}$ and $V = 20\ \text{V}$, so:

   $20 = \mathcal{E} - 2r \quad \text{(1)}$

   2. Reverse the current When the current direction is reversed, the cell is being charged: current enters its positive terminal. Now the internal resistance causes a voltage drop opposing the charging source, so the terminal voltage becomes:

   $V' = \mathcal{E} + I r$

   (The emf and the internal drop now add.)

   Given $V' = 30\ \text{V}$ with the same magnitude of current $2\ \text{A}$:

   $30 = \mathcal{E} + 2r \quad \text{(2)}$

   3. Solve the system Subtract equation (1) from equation (2): …

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10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.A solenoid of 1000 turns per metre has a core of material with relative permeability 400. The windings of the solenoid are insulated from the core and a current of 2 A is passed through the solenoid. Then the value of the magnetic intensity inside the solenoid is (A) $2 \times 10^3 \ \mathrm{Am}^{-1}$ (B) $1.0 \ \mathrm{Am}^{-1}$ (C) $8 \times 10^5 \ \mathrm{Am}^{-1}$ (D) $794 \ \mathrm{Am}^{-1}$

    ›Reveal solutionSolution

    Magnetic intensity $H$ depends only on the free current and geometry, not on the core material. For a solenoid, $H = nI$, giving $H = 2000\ \mathrm{A\,m^{-1}}$, so the correct option is (A).

    The key concept here is the distinction between magnetic intensity $\mathbf{H}$ (also called magnetizing field) and magnetic flux density $\mathbf{B}$. Many students mistakenly think that the core’s relative permeability $\mu_r$ affects $H$, but it does not. $H$ is produced solely by free currents (the current in the solenoid windings) and the geometry. The core material only determines how much $B$ results from that $H$ via $B = \mu_0 \mu_r H$.

    Let’s work through it step by step.

    1. Recall the definition of magnetic intensity for a solenoid For an ideal long solenoid, the magnetic intensity $H$ inside is uniform and given by

    $H = n I$

    where $n$ is the number of turns per unit length and $I$ is the current. This formula comes from Ampère’s law applied to the free current only: $\oint \mathbf{H} \cdot d\mathbf{l} = I_{\text{free, enclosed}}$.

    2. Plug in the given numbers Here, $n = 1000\ \text{turns per metre}$ and $I = 2\ \mathrm{A}$. So

    $H = 1000 \times 2 = 2000\ \mathrm{A\,m^{-1}}.$

    3. Notice that the core’s relative permeability $\mu_r = 400$ is irrelevant …

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11. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.A solenoid of 1000 turns per metre has a core of material with relative permeability 400. The windings of the solenoid are insulated from the core and a current of 2 A is passed through the solenoid. Then the value of the magnetic intensity inside the solenoid is (A) $2 \times 10^3~\mathrm{Am}^{-1}$ (B) $8 \times 10^5~\mathrm{Am}^{-1}$ (C) $1.0~\mathrm{Am}^{-1}$ (D) $794~\mathrm{Am}^{-1}$

    ›Reveal solutionSolution

    Magnetic intensity $H$ depends only on the free current and geometry, not on the core material. For a solenoid, $H = nI$. Here $n = 1000$ turns/m, $I = 2$ A, so $H = 2000\ \mathrm{A/m}$. The correct option is (A).

    Concept & Intuition

    The magnetic intensity $\mathbf{H}$ (also called the magnetizing field) is a field that accounts only for free currents (the current in the wires). Inside a solenoid, it is given by $H = n I$, where $n$ is the number of turns per unit length and $I$ is the current. Crucially, $H$ does not depend on the core material — the relative permeability $\mu_r$ affects the magnetic flux density $B = \mu_0 \mu_r H$, but not $H$ itself. Many students mistakenly plug $\mu_r$ into the formula for $H$, but that would be wrong: $H$ is defined by Ampere’s law for free currents alone.

    Step-by-step reasoning

    1. Identify the relevant formula For an ideal solenoid, Ampere’s law for the magnetic intensity $\mathbf{H}$ is

    $\oint \mathbf{H} \cdot d\mathbf{l} = I_{\text{free, enclosed}}$

    Inside a long solenoid, this gives

    $H = n I$

    where $n$ is the number of turns per unit length and $I$ is the current in each turn.

    2. Plug in the given numbers Here $n = 1000$ turns per metre and $I = 2$ A.

    $H = (1000\ \text{turns/m}) \times (2\ \text{A}) = 2000\ \mathrm{A/m}$

    3. Check the role of the core The relative permeability $\mu_r = 400$ is given, but it does not appear in the expression for $H$. It would be used to find the magnetic flux density: …

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12. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.At a certain place in the magnetic meridian, the earth's magnetic field is twice its vertical component. The ratio of horizontal component of earth's magnetic field and the total magnetic field of the earth at that place is (A) $\sqrt{3}:2$ (B) $1:2$ (C) $1:\sqrt{3}$ (D) $1:3$

    ›Reveal solutionSolution

    The Earth’s total magnetic field $B$ is related to its vertical component $B_V$ and horizontal component $B_H$ by $B^2 = B_H^2 + B_V^2$. Given $B = 2B_V$, we find $B_H = \sqrt{3}\,B_V$, so the ratio $B_H : B = \sqrt{3} : 2$. The correct option is (A).

    Concept & Intuition

    The Earth’s magnetic field at any point can be resolved into a horizontal component $B_H$ (parallel to the surface) and a vertical component $B_V$ (pointing into or out of the ground). These are perpendicular, so the total field $B$ is the hypotenuse of a right triangle:

    $B^2 = B_H^2 + B_V^2.$

    The problem tells us that at this location, the total field is twice the vertical component: $B = 2B_V$. That immediately gives a relationship between the sides, and we can solve for the ratio $B_H : B$.

    Step-by-step reasoning

    1. Write the given condition The Earth’s total magnetic field $B$ is twice its vertical component:

    $B = 2B_V.$

    2. Use the Pythagorean relation Since $B_H$ and $B_V$ are perpendicular:

    $B^2 = B_H^2 + B_V^2.$

    Substitute $B = 2B_V$:

    $(2B_V)^2 = B_H^2 + B_V^2.$

    3. Solve for $B_H$ in terms of $B_V$ $$$$

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