Q.A rectangular frame of wire is placed in a uniform magnetic field directed outwards, normal to the paper. AB is connected to a spring which is stretched to $A'B'$ and then released at time $t = 0$. Explain qualitatively how induced e.m.f. in the coil would vary with time. (Neglect damping of oscillations of spring)

Showing the 12 most recent of 26 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.The mean radius of a Rowland ring is 12 cm and it has 3000 turns of wire wound on its ferromagnetic core of relative permeability 500. If the magnetic field inside the core is $5\,\mathrm{T}$, then the magnetizing current is (A) $2\,\mathrm{A}$ (B) $3\,\mathrm{A}$ (C) $4\,\mathrm{A}$ (D) $5\,\mathrm{A}$

   ›Reveal solutionSolution

   The key is to relate the magnetic field $B$ to the magnetizing current $I$ using Ampere’s law for a toroidal core: $B = \mu_0 \mu_r \frac{N I}{2\pi r}$. Solving gives $I \approx 2\,\mathrm{A}$, so the correct option is (A).

   Concept & Intuition

   A Rowland ring is essentially a toroidal solenoid — a coil wound around a doughnut-shaped ferromagnetic core. The magnetic field inside such a core is not just due to the current in the wire; the core’s high relative permeability $\mu_r$ greatly amplifies the field. Ampere’s law for a toroid tells us that the line integral of the magnetic field $\mathbf{H}$ around a closed loop inside the core equals the total current threading the loop. Since the field is uniform along the circular path of radius $r$ (the mean radius), we get a simple algebraic relation. The trick is to remember that $B = \mu_0 \mu_r H$, so we first find $H$ from $B$ and $\mu_r$, then use Ampere’s law to find the current.

   Step-by-step solution

   1. Identify the geometry and relevant law The mean radius of the ring is $r = 12\,\text{cm} = 0.12\,\text{m}$. The number of turns is $N = 3000$. The relative permeability is $\mu_r = 500$, and the magnetic field inside the core is $B = 5\,\text{T}$. For a toroid, Ampere’s law states:

   $\oint \mathbf{H} \cdot d\mathbf{l} = N I$

   where $I$ is the current in the wire. Because the field is constant along the circular path of circumference $2\pi r$, this becomes:

   $H \cdot (2\pi r) = N I$

   2. Relate $H$ to $B$ using the core’s permeability In a magnetic material, $B = \mu_0 \mu_r H$, so:

   $H = \frac{B}{\mu_0 \mu_r}$

   Here $\mu_0 = 4\pi \times 10^{-7}\,\text{T·m/A}$.

   3. Substitute $H$ into Ampere’s law From step 1: $I = \frac{H \cdot 2\pi r}{N}$. Replace $H$:

   $I = \frac{2\pi r}{N} \cdot \frac{B}{\mu_0 \mu_r}$

   4. Plug in the numbers

   $I = \frac{2\pi (0.12)}{3000} \cdot \frac{5}{(4\pi \times 10^{-7}) \cdot 500}$

   Simplify step by step:

   • $2\pi \times 0.12 = 0.24\pi$
   • Denominator of the first fraction: $3000$
   • Second fraction: $\frac{5}{4\pi \times 10^{-7} \times 500} = \frac{5}{2\pi \times 10^{-4}} = \frac{5}{2\pi} \times 10^4$ So:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the current passing through a coil of self-inductance 50 mH having 50 turns is 5 A, then the magnetic flux linked with the coil (in $10^{-3}$ Wb) is (A) 5 (B) 20 (C) 10 (D) 250

   ›Reveal solutionSolution

   The magnetic flux linked with a coil is directly proportional to the current flowing through it, with the constant of proportionality being the self-inductance. Using the given self-inductance and current, the total magnetic flux linked with the coil is $\boxed{250 \times 10^{-3} \text{ Wb}}$.

   Concept and Intuition

   When an electric current flows through a coil, it generates a magnetic field. This magnetic field, in turn, produces magnetic flux that passes through the coil itself. The key idea here is that the magnetic flux linked with the coil is directly proportional to the current flowing through it. This proportionality is quantified by a property of the coil called self-inductance.

   The total magnetic flux ($\Phi_B$) linked with a coil is given by:

   $\Phi_B = L I$

   where $L$ is the self-inductance of the coil and $I$ is the current flowing through it.

   It's important to understand what "magnetic flux linked with the coil" means. For a coil with $N$ turns, if $\phi$ is the magnetic flux passing through a single turn, then the total magnetic flux linked with the coil is $N\phi$. The self-inductance $L$ is defined such that $L = \frac{N\phi}{I}$. Therefore, when we use the formula $\Phi_B = L I$, the $\Phi_B$ here already represents the total flux linkage ($N\phi$). The number of turns ($N$) is already factored into the value of $L$ itself. This means we do not need to multiply by the number of turns again if $L$ is provided directly.

   Step-by-Step Solution

   1. Identify the given values:

      • Self-inductance of the coil, $L = 50 \text{ mH}$
      • Current flowing through the coil, $I = 5 \text{ A}$
      • Number of turns, $N = 50 \text{ turns}$ (As explained above, this information is implicitly included in the value of $L$ and is not needed for the calculation when $L$ is given directly.)

   2. Convert units to SI base units:

      The self-inductance is given in millihenries (mH). We need to convert it to Henries (H) for consistency in SI units.

      $1 \text{ mH} = 10^{-3} \text{ H}$

      So, $L = 50 \text{ mH} = 50 \times 10^{-3} \text{ H}$.

   3. Apply the formula for magnetic flux linkage: …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The frequency of an alternating voltage is 50 Hz. The time taken for instantaneous voltage to increase from zero to half of its peak voltage is (A) $\dfrac{1}{800}$ s (B) $\dfrac{1}{600}$ s (C) $\dfrac{1}{300}$ s (D) $\dfrac{1}{200}$ s

   ›Reveal solutionSolution

   For a sinusoidal voltage, the time to rise from zero to half the peak is one‑twelfth of a period. With frequency 50 Hz (period 1/50 s), that time is 1/600 s → option (B).

   Concept & Intuition

   An alternating voltage is usually sinusoidal: $v(t) = V_0 \sin(\omega t)$, where $V_0$ is the peak voltage and $\omega = 2\pi f$. The question asks for the first time after $t=0$ that the voltage reaches $V_0/2$. On a sine wave, the value $\frac12$ occurs when the angle is $\pi/6$ (30°) because $\sin(\pi/6)=0.5$. So we just need to find how long it takes for the phase $\omega t$ to become $\pi/6$. That’s a fraction of the full cycle (2π rad), and the period is $T = 1/f$.

   ---

   Step‑by‑step solution

   1. Write the voltage equation

   $v(t) = V_0 \sin(2\pi f t)$

   with $f = 50\ \text{Hz}$.

   2. Set the condition We want $v(t) = \frac{V_0}{2}$, so

   $V_0 \sin(2\pi f t) = \frac{V_0}{2} \quad\Rightarrow\quad \sin(2\pi f t) = \frac12.$

   3. Find the smallest positive angle The sine equals $\frac12$ at angles $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ (and then every $2\pi$). The first time after zero is the smallest positive angle:

   $2\pi f t = \frac{\pi}{6}.$

   4. Solve for time $t = \frac{\pi/6}{2\pi f} = \frac{1}{12 f}.$ …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A straight uniform wire of resistance $36~\Omega$ is bent in the form of a semi-circular loop. The effective resistance between the ends of the diameter of the semi-circular loop is (A) $\dfrac{36}{7}~\Omega$ (B) $\dfrac{77}{9}~\Omega$ (C) $\dfrac{99}{7}~\Omega$ (D) $\dfrac{56}{9}~\Omega$

   ›Reveal solutionSolution

   The wire is bent into a semicircle, so the total resistance is split into two arcs in parallel; the effective resistance between the diameter ends is $9\ \Omega$, which corresponds to option (D).

   Concept & Intuition

   When a uniform wire is bent into a shape, the resistance of any segment is proportional to its length. Here, the wire is bent into a semicircular loop. The two ends of the diameter are the endpoints of the semicircle. The current can travel from one end to the other along two paths: the upper arc and the lower arc. These two arcs are connected in parallel between the same two points. So the problem reduces to finding the resistances of the two arcs and then combining them in parallel.

   ---

   1. Total resistance and total length

      The straight wire has resistance $R_{\text{total}} = 36\ \Omega$. When bent into a semicircle, the total length of the wire is unchanged. Let the length of the whole wire be $L$. Then the resistance per unit length is $\frac{36}{L}\ \Omega/\text{unit length}$.

   2. Shape of the semicircular loop

      A semicircular loop has two arcs: the semicircular arc itself (the curved part) and the straight diameter. But careful — the problem says the wire is bent into a semi-circular loop. That means the wire forms the curved part only; the diameter is not part of the wire. The ends of the diameter are the two endpoints of the semicircular arc. So the wire is just the curved semicircle.

   3. Lengths of the two paths

      Let the radius of the semicircle be $r$.

      • The curved arc (the wire itself) has length $\pi r$.
      • The straight-line distance between the two ends is the diameter, length $2r$. But that straight line is not a wire — it's just the gap. So the only conducting paths are along the wire itself. However, between the two ends, the wire forms two arcs? No — a semicircle has only one arc. Wait, this is the key subtlety.

      Actually, when you bend a straight wire into a semicircle, the two ends are the endpoints of the semicircle. The wire itself is a single continuous arc. So between the two ends, there is exactly one conducting path: the semicircular arc. That would give resistance simply $36\ \Omega$. But that's not among the options, so the interpretation must be different.

   4. Correct interpretation

      The phrase "bent in the form of a semi-circular loop" often means the wire is shaped into a closed semicircle — that is, the wire forms both the curved part and the diameter. In other words, the wire is bent into a semicircular loop (a closed shape), so the wire itself includes the straight diameter segment. Then the total length is the sum of the curved part and the diameter.

      Let the radius be $r$.

      • Curved arc length = $\pi r$
      • Diameter length = $2r$
      • Total length = $\pi r + 2r = r(\pi + 2)$

      Since total resistance is $36\ \Omega$, resistance per unit length = $\frac{36}{r(\pi+2)}$.

   5. Resistances of the two paths between diameter ends

      The two ends of the diameter are the endpoints of both the curved arc and the diameter itself. So between these two points, there are two parallel paths:

      • Path 1: the curved arc, length $\pi r$, resistance $R_1 = \frac{36}{r(\pi+2)} \cdot \pi r = \frac{36\pi}{\pi+2}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.A wheel with 12 metallic spokes each 40 cm long is rotated with an angular speed of 15 rad s$^{-1}$ in a plane normal to the horizontal component of earth’s magnetic field. If the horizontal component of earth’s magnetic field at the place is $4 \times 10^{-5}$ T, then the induced emf between the axle and the rim of the wheel is (A) $1.2 \times 10^{-5}$ V (B) $4.8 \times 10^{-5}$ V (C) $2.4 \times 10^{-5}$ V (D) $3.6 \times 10^{-5}$ V

   ›Reveal solutionSolution

   Each spoke acts as a rotating rod cutting the Earth’s horizontal magnetic field. The induced emf between the axle and rim is the same as that from one spoke, giving $4.8 \times 10^{-5}$ V.

   The key idea: when a conductor rotates in a uniform magnetic field, an emf is induced across its ends because each point on the conductor moves with a different velocity. For a spoke, one end (the axle) is stationary, and the other end (the rim) moves with the maximum speed. The emf across a rotating rod is found by integrating the motional emf along its length.

   Since all 12 spokes are connected in parallel between the axle and the rim, the net emf is just the emf of a single spoke — not 12 times that. This is a classic trap.

   1. Identify the relevant formula. For a rod of length $L$ rotating about one end with angular speed $\omega$ in a uniform magnetic field $B$ perpendicular to the plane of rotation, the induced emf is

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

   Why? Consider a small element $dr$ at distance $r$ from the axle. Its linear velocity is $v = \omega r$, perpendicular to both $B$ and $dr$. The motional emf across that element is $d\mathcal{E} = B v \, dr = B \omega r \, dr$. Integrate from $r=0$ to $r=L$:

   $\mathcal{E} = \int_0^L B \omega r \, dr = \frac{1}{2} B \omega L^2.$

   2. Plug in the given values. $B = 4 \times 10^{-5}$ T, $\omega = 15$ rad/s, $L = 40$ cm $= 0.4$ m.

   $\mathcal{E} = \frac{1}{2} \times (4 \times 10^{-5}) \times 15 \times (0.4)^2.$

   3. Calculate step by step. $(0.4)^2 = 0.16$. $15 \times 0.16 = 2.4$. …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.The magnetic field (in $10^{-4}$ T) at the centre of a toroid of mean radius 10 cm with 200 turns and carrying a current of 2.5 A is (A) $2\pi$ (B) $10$ (C) $5$ (D) Zero

   ›Reveal solutionSolution

   The magnetic field at the centre (central hole) of a toroid is zero.

   A toroid is a solenoid bent into a closed ring. Applying Ampere's law to a circular Amperian loop:

   • Inside the windings (through the core), B = mu0 N I / (2 pi r), non-zero.
   • In the central hollow region (the geometric centre of the toroid) and outside the windings, the net current enclosed by an Amperian loop is zero, so B = 0. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.A train is moving with a speed of 198 kmph. If the distance between the rails is 120 cm and the vertical component of earth's magnetic field is $0.25 \times 10^{-4}$ T, the induced emf between the ends of the axle of the train is (A) 1.95 mV (B) 2.2 mV (C) 1.65 mV (D) 3.3 mV

   ›Reveal solutionSolution

   A metal axle moving perpendicular to the vertical magnetic field acts like a moving rod in a field; the induced emf is $\mathcal{E} = B v L$, giving 1.65 mV.

   The key idea is motional emf. When a conductor moves through a magnetic field, the free electrons inside experience a magnetic force, which pushes them to one end until the resulting electric field balances it. The potential difference that develops across the ends is the induced emf. Here the axle of the train is a metal rod of length equal to the distance between the rails, moving horizontally. The vertical component of Earth’s field is the one that cuts the axle, so only that component matters.

   Let’s work it through.

   1. Convert all quantities to SI units.

      Speed: $198 \text{ km/h} = 198 \times \frac{1000}{3600} = 55 \text{ m/s}$.

      Axle length (distance between rails): $120 \text{ cm} = 1.2 \text{ m}$.

      Vertical magnetic field: $B_v = 0.25 \times 10^{-4} \text{ T}$.

   2. Apply the motional emf formula.

      For a rod of length $L$ moving with velocity $v$ perpendicular to a uniform magnetic field $B$, the induced emf is

   $\mathcal{E} = B v L.$

   Here the motion is horizontal and the field is vertical, so they are perpendicular — exactly the situation the formula describes.

   3. Plug in the numbers.

   $\mathcal{E} = (0.25 \times 10^{-4}) \times 55 \times 1.2.$

   First multiply $55 \times 1.2 = 66$.

   Then $0.25 \times 10^{-4} \times 66 = 16.5 \times 10^{-4} = 1.65 \times 10^{-3} \text{ V}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.A train with an axle of length $1.66$ m is moving towards north with a speed of $90$ kmh$^{-1}$. If the vertical component of the earth’s magnetic field is $0.2 \times 10^{-4}$ T, the emf induced across the ends of the axle of the train is (A) $16.6$ mV (B) $1.66$ mV (C) $0.83$ mV (D) $8.3$ mV

   ›Reveal solutionSolution

   The induced emf is given by motional emf formula $\mathcal{E} = B v L$, where $B$ is the vertical component of Earth’s magnetic field, $v$ is the train’s speed, and $L$ is the axle length. After converting units, the result is $0.83$ mV, so the correct option is (C).

   The key concept here is motional emf. When a conductor moves through a magnetic field, the free charges inside experience a magnetic force, causing them to separate until an electric field balances it. This separation creates a potential difference — the induced emf. For a straight conductor of length $L$ moving with velocity $v$ perpendicular to a uniform magnetic field $B$, the induced emf is simply $\mathcal{E} = B v L$. In this problem, the axle is horizontal and moving north, while the vertical component of Earth’s field points downward (or upward) — crucially, the motion is perpendicular to the vertical field component, so the formula applies directly.

   Let’s work through it step by step.

   1. Identify the given quantities

      • Axle length: $L = 1.66$ m
      • Train speed: $v = 90$ km/h
      • Vertical component of Earth’s magnetic field: $B = 0.2 \times 10^{-4}$ T

   2. Convert speed to SI units

      The formula requires speed in m/s.

   $v = 90 \ \frac{\text{km}}{\text{h}} = 90 \times \frac{1000}{3600} \ \text{m/s} = 90 \times \frac{5}{18} = 25 \ \text{m/s}$

   (A handy conversion: multiply km/h by $\frac{5}{18}$ to get m/s.)

   3. Apply the motional emf formula

   $\mathcal{E} = B v L$

   Substitute the values:

   $\mathcal{E} = (0.2 \times 10^{-4}) \times 25 \times 1.66$

   4. Calculate step by step First, $0.2 \times 10^{-4} = 2 \times 10^{-5}$ T. …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.A coil of 100 turns and $0.10\,\mathrm{m}^2$ area, making two rotations per second is placed in a $0.01\,\mathrm{T}$ uniform magnetic field perpendicular to its axis of rotation. The maximum voltage generated in the coil is (A) $0.1\,\mathrm{V}$ (B) $12.56\,\mathrm{V}$ (C) $1.256\,\mathrm{V}$ (D) $0.628\,\mathrm{V}$

   ›Reveal solutionSolution

   The maximum voltage generated in a coil rotating in a uniform magnetic field is determined by the rate of change of magnetic flux through it. Using Faraday's Law, the maximum induced EMF is $NBA\omega$, which calculates to $\boxed{1.256\,\mathrm{V}}$.

   When a coil rotates in a uniform magnetic field, the magnetic flux passing through it changes continuously. This change in magnetic flux induces an electromotive force (EMF), or voltage, in the coil, as described by Faraday's Law of Electromagnetic Induction. The induced EMF is proportional to the rate at which the magnetic flux changes.

   For a coil rotating with a constant angular velocity, the magnetic flux varies sinusoidally with time. Consequently, the induced EMF also varies sinusoidally. The maximum EMF is generated when the rate of change of magnetic flux is highest, which occurs when the plane of the coil is parallel to the magnetic field lines.

   The instantaneous magnetic flux $\Phi$ through a coil of $N$ turns, area $A$, rotating with angular velocity $\omega$ in a magnetic field $B$ is given by:

   $\Phi = NBA \cos(\omega t)$

   The induced EMF $E$ is given by Faraday's Law:

   $E = -\frac{d\Phi}{dt}$

   Let's apply these concepts to solve the problem.

   1. Identify the given parameters:

      • Number of turns, $N = 100$
      • Area of the coil, $A = 0.10\,\mathrm{m}^2$
      • Frequency of rotation, $f = 2\,\mathrm{rotations/second}$
      • Magnetic field strength, $B = 0.01\,\mathrm{T}$

   2. Calculate the angular velocity ($\omega$):

      The frequency $f$ is given in rotations per second. Angular velocity $\omega$ is related to frequency by the formula $\omega = 2\pi f$.

   $\omega = 2\pi (2\,\mathrm{s}^{-1}) = 4\pi\,\mathrm{rad/s}$

   3. Determine the instantaneous magnetic flux ($\Phi$): The magnetic flux through the coil at any instant $t$ is given by $\Phi = NBA \cos(\theta)$, where $\theta$ is the angle between the magnetic field vector and the normal to the coil's plane. Since the coil rotates with angular velocity $\omega$, this angle can be expressed as $\theta = \omega t$ (assuming $\theta=0$ at $t=0$).

   $\Phi = NBA \cos(\omega t)$

   Substituting the given values:
   

   $\Phi = (100)(0.01\,\mathrm{T})(0.10\,\mathrm{m}^2) \cos(4\pi t)$

   $\Phi = 0.1 \cos(4\pi t)\,\mathrm{Wb}$

   4. Calculate the induced EMF ($E$) using Faraday's Law: Faraday's Law states that the induced EMF is the negative rate of change of magnetic flux with respect to time: $E = -\frac{d\Phi}{dt}$.

   $E = -\frac{d}{dt} [NBA \cos(\omega t)]$

   $E = -NBA (-\sin(\omega t)) (\omega)$

   $$ E = NBA\omega \sin(\omega t) $$ …
   

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.An arc making an angle $15^\circ$ at the center is removed from a ring of mass ‘M’ and radius ‘R’. The moment of inertia of the remaining ring about an axis passing through its center and perpendicular to its plane is (A) $\dfrac{23\,MR^2}{24}$ (B) $\dfrac{MR^2}{24}$ (C) $\dfrac{MR^2}{23}$ (D) $\dfrac{24\,MR^2}{23}$

    ›Reveal solutionSolution

    The moment of inertia of a complete ring is $MR^2$; removing a $15^\circ$ arc (which is $1/24$ of the ring) reduces the mass proportionally, and because every piece of the ring is at the same radius $R$, the remaining inertia is simply the full inertia minus the inertia of the removed piece: $\frac{23}{24}MR^2$. The correct option is (A).

    ---

    The key idea is that for a thin ring, every infinitesimal mass element lies at the same distance $R$ from the center. The moment of inertia about the central perpendicular axis is therefore just $I = M R^2$ for the whole ring. When we remove a portion, we are simply subtracting the contribution of that portion — no parallel axis theorem or integration over varying distances is needed.

    A common pitfall is to think the missing arc changes the shape in a way that alters the radius for some parts, but it doesn’t: the remaining pieces are still at radius $R$. So the problem reduces to a simple proportion.

    ---

    1. Find the fraction of the ring removed. The full ring subtends $360^\circ$. The removed arc subtends $15^\circ$.

    $\text{Fraction removed} = \frac{15^\circ}{360^\circ} = \frac{1}{24}.$

    2. Mass of the removed arc. The ring has uniform mass per unit length. So the mass of the removed arc is

    $m_{\text{removed}} = \frac{1}{24} M.$

    3. Moment of inertia of the removed arc. Every bit of the removed arc is at distance $R$ from the center. Its moment of inertia is $I_{\text{removed}} = m_{\text{removed}} R^2 = \frac{1}{24} M R^2.$ …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.A circular coil of area $200\ \text{cm}^2$ and $50$ turns is rotating about its vertical diameter with an angular speed of $40\ \text{rads}^{-1}$ in a uniform horizontal magnetic field of magnitude $2 \times 10^{-2}\ \text{T}$. The maximum emf induced in the coil is (A) $1.2\ \text{V}$ (B) $0.8\ \text{V}$ (C) $0.6\ \text{V}$ (D) $0.3\ \text{V}$

    ›Reveal solutionSolution

    The maximum induced emf in a rotating coil in a uniform magnetic field is given by $\mathcal{E}_{\text{max}} = N B A \omega$. Substituting the given values yields $0.8\ \text{V}$, so the correct option is (B).

    Concept & Intuition

    When a coil rotates in a magnetic field, the magnetic flux through it changes sinusoidally with time. Faraday’s law tells us that the induced emf is the rate of change of flux. The maximum emf occurs when the plane of the coil is parallel to the field (flux is zero but changing fastest). The formula $\mathcal{E}_{\text{max}} = N B A \omega$ comes directly from differentiating $\Phi = N B A \cos(\omega t)$.

    Step-by-step solution

    1. Identify the given quantities

       • Area of coil: $A = 200\ \text{cm}^2 = 200 \times 10^{-4}\ \text{m}^2 = 0.02\ \text{m}^2$
       • Number of turns: $N = 50$
       • Angular speed: $\omega = 40\ \text{rad/s}$
       • Magnetic field: $B = 2 \times 10^{-2}\ \text{T}$

    2. Recall the formula for maximum induced emf

       For a coil rotating in a uniform magnetic field, the flux at time $t$ is $\Phi = N B A \cos(\omega t)$.

       The induced emf is $\mathcal{E} = -\frac{d\Phi}{dt} = N B A \omega \sin(\omega t)$.

       The maximum value occurs when $\sin(\omega t) = 1$, so

    $\mathcal{E}_{\text{max}} = N B A \omega.$

    3. Substitute the values

    $\mathcal{E}_{\text{max}} = 50 \times (2 \times 10^{-2}) \times 0.02 \times 40$

    Compute step by step:

    • $50 \times 2 \times 10^{-2} = 100 \times 10^{-2} = 1$
    • $1 \times 0.02 = 0.02$ …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
    Q.A straight conductor of length 150 cm moves with a velocity of $8~\mathrm{ms}^{-1}$ perpendicular to a magnetic field. If the induced emf across the ends of the conductor is $3~\mathrm{V}$, the magnitude of the magnetic field is (A) $1.5~\mathrm{T}$ (B) $0.75~\mathrm{T}$ (C) $0.5~\mathrm{T}$ (D) $0.25~\mathrm{T}$

    ›Reveal solutionSolution

    The induced emf in a moving conductor is given by $\mathcal{E} = B l v$ when $l$, $v$, and $B$ are mutually perpendicular. Using the given values, $B = \frac{3}{1.5 \times 8} = 0.25~\mathrm{T}$, so the correct option is (D).

    The core idea here is motional emf. When a conductor moves through a magnetic field, the free electrons inside experience a magnetic force, which pushes them to one end. This separation of charge creates an electric field, and the potential difference that builds up across the ends is the induced emf. The formula $\mathcal{E} = B l v$ applies only when the velocity, the magnetic field, and the length of the conductor are all mutually perpendicular — which is exactly the case given in the problem.

    Let’s work through it step by step.

    1. Identify the known quantities.

       Length of the conductor, $l = 150~\mathrm{cm} = 1.5~\mathrm{m}$ (always convert to SI units).

       Velocity, $v = 8~\mathrm{m/s}$.

       Induced emf, $\mathcal{E} = 3~\mathrm{V}$.

       The motion is perpendicular to the magnetic field, and the conductor is straight, so $l$, $v$, and $B$ are all at right angles to each other.

    2. Write the formula for motional emf.

       For perpendicular motion, the magnitude is

    $\mathcal{E} = B l v.$

    This comes from the Lorentz force: the magnetic force per unit charge is $vB$, and when this acts over the length $l$, the work done per unit charge (which is the emf) is $vBl$.

    3. Rearrange to solve for $B$.

    $B = \frac{\mathcal{E}}{l v}.$

    4. Substitute the values. …

    🔒 Start your 14-day free trial to see the full solution →