આ MCQ મોડ્યુલ આના પર આધારિત છે: NCERT Exercises and Solutions: Electromagnetic Induction
આ મૂલ્યાંકન આના પર આધારિત હશે: NCERT Exercises and Solutions: Electromagnetic Induction
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
This chapter showed how a changing magnetic flux generates electricity - EMI - and how this principle is harnessed in motors, generators, transformers and a host of modern devices.
| Term | Definition / Formula |
|---|---|
| Magnetic flux ΦB | ∫ B·dA; weber (Wb) = T·m². |
| EMF ε | Work done per unit charge by the source; volt (V). |
| Faraday's law | ε = −dΦ/dt; ε = −N dΦ/dt for N-turn coil. |
| Lenz's law | Induced current opposes the change in flux. |
| Motional EMF | ε = BLv; rod moving perpendicular to B. |
| Eddy current | Loops of induced current within a conductor's body. |
| Self-inductance L | NΦ = LI. Solenoid: L = μ₀ N²A/ℓ. Henry (H). |
| Mutual inductance M | N₂Φ₂ = M I₁. Coaxial solenoids: M = μ₀N₁N₂A₁/ℓ. |
| Energy in inductor | U = ½ L I². |
| Peak EMF (AC gen.) | ε₀ = NBAω. |
Predict the direction of induced current in the situations described by the following figures (a-f) [described verbally below]:
(a) A rectangular loop with bar magnet's S-pole moving toward it.
(b) A bar magnet moved away from a circular loop, N-pole facing loop.
(c) A rod moving on a U-rail to the right in field B (out of page).
(d) Decreasing current in a primary coil linked to a secondary loop.
Apply Lenz's law in each case.
(a) S-pole approaches ⇒ flux into the loop increases ⇒ induced current makes the near face S (so anti-clockwise as seen from the magnet, to repel).
(b) N-pole moving away ⇒ flux through loop (toward magnet) decreases ⇒ induced current tries to maintain it ⇒ near face becomes S (anti-clockwise viewed from magnet).
(c) Area enclosed grows ⇒ flux out of page increases ⇒ induced current circulates clockwise (as seen by viewer) to create opposing flux into page.
(d) Decreasing primary current ⇒ flux through secondary decreases ⇒ secondary current circulates to maintain it ⇒ in the same sense as primary current.
Use Lenz's law to determine the direction of induced current in the situations described by figure (irregular loop becoming circular while in a uniform B; circular loop being deformed into a narrow shape).
(a) An irregular loop becoming circular: maximum area is the circle. If B is into the page, area increases ⇒ flux into page increases ⇒ induced current is anti-clockwise (creates flux out of page).
(b) Circular loop deformed into a narrow shape: area decreases ⇒ flux into page decreases ⇒ induced current is clockwise (maintains flux into page).
A long solenoid with 15 turns/cm has a small loop of area 2.0 cm² placed inside it normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced EMF in the loop while the current is changing?
n = 15 turns/cm = 1500 turns/m; B inside = μ₀nI.
dB/dt = μ₀n(dI/dt) = (4π×10⁻⁷)(1500)(2/0.1) = (4π×10⁻⁷)(1500)(20) = 3.77 × 10⁻² T/s.
ε = (dB/dt) × A = 3.77 × 10⁻² × 2.0 × 10⁻⁴ = 7.54 × 10⁻⁶ V ≈ 7.5 μV.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the EMF developed across the cut if the velocity of the loop is 1 cm/s in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
(a) Moving normal to longer side (8 cm): the leaving edge has length L = 2 cm = 0.02 m. ε = BLv = 0.3 × 0.02 × 0.01 = 6 × 10⁻⁵ V = 60 μV. Time = 8 cm / 1 cm/s = 8 s.
(b) Moving normal to shorter side (2 cm): leaving edge has length L = 8 cm = 0.08 m. ε = 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴ V = 240 μV. Time = 2 s.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the EMF developed between the centre and the ring.
For a rotating rod: ε = ½ B ω L² = 0.5 × 0.5 × 400 × (1.0)² = 100 V.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the Earth's magnetic field, 0.30 × 10⁻⁴ Wb/m². (a) Find the instantaneous value of the EMF induced in the wire. (b) What is the direction of the EMF? (c) Which end of the wire is at the higher potential?
(a) ε = BLv = 0.30 × 10⁻⁴ × 10 × 5 = 1.5 × 10⁻³ V = 1.5 mV.
(b) Force on free electrons F = −e (v × B). v points down; B (horizontal component) points north. v × B points west; force on electrons (negative charge) points east. Hence electrons accumulate at the east end, conventional current is east → west. Electrons accumulate at east; conventional current flows west to east inside wire.
(c) The west end is at higher potential.
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average EMF of 200 V is induced, give an estimate of the self-inductance of the circuit.
|ε| = L|dI/dt| ⇒ L = ε × Δt/ΔI = 200 × 0.1/5.0 = 4.0 H.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
N₂Φ₂ = M I₁ ⇒ change in flux linkage = M × ΔI = 1.5 × 20 = 30 Wb.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth's magnetic field at the location has a magnitude of 5 × 10⁻⁴ T and the dip angle is 30°?
v = 1800 km/h = 500 m/s. Vertical component of B: Bv = B sin 30° = 5×10⁻⁴ × 0.5 = 2.5 × 10⁻⁴ T (this is what cuts the horizontal wing).
ε = Bv L v = 2.5 × 10⁻⁴ × 25 × 500 = 3.125 V ≈ 3.13 V.
Suppose the loop in Exercise 6.4(a) is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T/s. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat?
ε = (dB/dt) × A = 0.02 × (0.08 × 0.02) = 0.02 × 1.6 × 10⁻³ = 3.2 × 10⁻⁵ V.
P = ε²/R = (3.2 × 10⁻⁵)² / 1.6 = 1.024 × 10⁻⁹ / 1.6 ≈ 6.4 × 10⁻¹⁰ W.
A square loop of side 12 cm with its sides parallel to x and y axes is moved with a velocity of 8 cm/s in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10⁻³ T/cm along the negative x-direction (that is, decreases by 10⁻³ T/cm as x increases) and decreases at the rate of 10⁻³ T/s. Find the direction and magnitude of the induced current in the loop if its resistance is 4.5 mΩ.
Two contributions: (i) loop moving into a region of decreasing B (gradient): EMF = (dB/dx) v × A = 10⁻³ × (1/10⁻²) × 0.08 × (0.12)² = 10⁻¹ × 0.08 × 0.0144 = 1.152 × 10⁻⁴ V (loop sees decreasing B).
Wait - convert gradient: 10⁻³ T/cm = 10⁻¹ T/m. Effective EMF from spatial gradient as loop moves: ε₁ = (dB/dx)(v) × A = (10⁻¹)(0.08)(0.12 × 0.12) = 1.152 × 10⁻⁴ V.
(ii) Time variation: ε₂ = (dB/dt)A = 10⁻³ × 0.0144 = 1.44 × 10⁻⁵ V.
Total EMF (both reduce flux into z, so both drive current in same sense, anti-clockwise when viewed from +z): ε = ε₁ + ε₂ = 1.152 × 10⁻⁴ + 1.44 × 10⁻⁵ ≈ 1.30 × 10⁻⁴ V.
I = ε/R = 1.30 × 10⁻⁴ / 4.5 × 10⁻³ ≈ 2.88 × 10⁻² A ≈ 28.9 mA, anti-clockwise as seen from +z.
A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis. A uniform magnetic field B extends over a circular region of radius a (with a < R), and is suddenly switched off. What is the angular velocity acquired by the wheel?
When B is switched off, induced electric field E_φ along rim is given by Faraday: ∮E·dl = −dΦ/dt; over a circle of radius R: 2πRE = −d(Bπa²)/dt ⇒ E = −(a²/2R)(dB/dt).
Force per unit length on rim: F = λE; total tangential force = λE × 2πR; torque τ = (λE × 2πR) × R = 2πλR²E = 2πλR² × (−a²/2R)(dB/dt) = −πλRa²(dB/dt).
Angular impulse: ΔL = ∫τ dt = −πλRa² ΔB = πλRa²B (since B decreases from B to 0).
L_final = MR² ω ⇒ ω = πλRa²B/(MR²) = πλa²B/(MR). Direction depends on sign of λ.
A 100-turn coil of area 0.10 m² rotates in a horizontal plane about a vertical axis at the rate of 0.5 rev/s in the horizontal component of Earth's magnetic field 7.0 × 10⁻⁵ T. Calculate the maximum and average EMF induced in the coil and the maximum current.
ω = 2π × 0.5 = π rad/s. ε₀ = NBAω = 100 × 7.0 × 10⁻⁵ × 0.10 × π ≈ 2.2 × 10⁻³ V = 2.2 mV.
Average EMF over a half-cycle = (2/π)ε₀ ≈ 1.4 mV; over a full cycle = 0.
Toggle modes - flux, motional, AC generator - and tune parameters. Outputs all updates live.
Options: (A) Both true, R correct explanation. (B) Both true, R not the explanation. (C) A true, R false. (D) A false, R true.
તમે જે ભણ્યા તેનું પૂરું પેપર આપો, પ્રશ્ન દીઠ તપાસાયેલું.
Section A · Section B · Section C · Section D · Section E