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Electromagnetic Induction - The Experiments of Faraday and Henry

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electromagnetic Induction is the phenomenon of generating an electromotive force (EMF) and an induced current by varying the magnetic flux linked with a conductor.

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Faraday and Henry's Experiment 1 (Magnet-Coil): When a bar magnet is moved towards or away from a stationary conducting coil, a deflection is observed in the galvanometer connected to the coil, indicating an induced current. The deflection lasts only as long as the magnet is in motion.

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Faraday and Henry's Experiment 2 (Coil-Coil): Replacing the magnet with a second coil carrying a steady current (acting as a primary coil) produces similar results. Relative motion between the primary coil and the secondary coil induces a current in the secondary coil.

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Faraday and Henry's Experiment 3 (Current Variation): Two stationary coils are placed near each other. When the current in the primary coil is switched 'ON' or 'OFF' using a tap key, a momentary deflection is observed in the galvanometer of the secondary coil. This proves that relative motion is not the only way to induce current; a change in magnetic field strength also works.

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Magnetic Flux (ΦB\Phi_B): It is the total number of magnetic field lines crossing a surface. It is a scalar quantity defined as the dot product of magnetic field B\mathbf{B} and area vector A\mathbf{A}.

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The SI unit of magnetic flux is the Weber (WbWb). 1 Wb=1 T⋅m21\, Wb = 1\, T \cdot m^2.

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The core conclusion of these experiments is that the change in magnetic flux linked with a coil is the fundamental cause of induced EMF.

📐Formulae

ΦB=B⋅A=BAcos⁡θ\Phi_B = \mathbf{B} \cdot \mathbf{A} = BA \cos \theta

ΔΦB=Φfinal−Φinitial\Delta \Phi_B = \Phi_{final} - \Phi_{initial}

[ΦB]=M1L2T−2A−1 (Dimensional Formula)[\Phi_B] = M^1 L^2 T^{-2} A^{-1} \text{ (Dimensional Formula)}

💡Examples

Problem 1:

A circular loop of radius r=7 cmr = 7 \text{ cm} is placed in a uniform magnetic field of B=0.5 TB = 0.5 \text{ T}. If the plane of the loop makes an angle of 30∘30^\circ with the magnetic field lines, calculate the magnetic flux through the loop. (Take π=227\pi = \frac{22}{7})

Solution:

  1. Area of the loop A=πr2=227×(0.07 m)2=227×0.0049=0.0154 m2A = \pi r^2 = \frac{22}{7} \times (0.07\, m)^2 = \frac{22}{7} \times 0.0049 = 0.0154\, m^2.
  2. The angle θ\theta is the angle between the Normal to the plane and the Magnetic Field. Given the plane makes 30∘30^\circ with BB, the normal makes θ=90∘−30∘=60∘\theta = 90^\circ - 30^\circ = 60^\circ with BB.
  3. Calculate flux using ΦB=BAcos⁡θ\Phi_B = BA \cos \theta: ΦB=0.5×0.0154×cos⁡60∘\Phi_B = 0.5 \times 0.0154 \times \cos 60^\circ ΦB=0.5×0.0154×0.5=0.00385 Wb=3.85×10−3 Wb\Phi_B = 0.5 \times 0.0154 \times 0.5 = 0.00385\, Wb = 3.85 \times 10^{-3}\, Wb.

Explanation:

Magnetic flux depends on the orientation of the area vector. Since the area vector is perpendicular to the plane of the coil, the angle used in the cosine function must be the complement of the angle given relative to the plane.

Problem 2:

During an experiment, the magnetic flux through a coil changes from 1.2×10−2 Wb1.2 \times 10^{-2}\, Wb to 0.4×10−2 Wb0.4 \times 10^{-2}\, Wb. What is the change in flux?

Solution:

0.012−0.0040.008\begin{array}{r} 0.012 \\ - 0.004 \\ \hline 0.008 \end{array} Change in flux ΔΦB=Φ2−Φ1=0.4×10−2−1.2×10−2=−0.8×10−2 Wb\Delta \Phi_B = \Phi_2 - \Phi_1 = 0.4 \times 10^{-2} - 1.2 \times 10^{-2} = -0.8 \times 10^{-2}\, Wb.

Explanation:

The change in flux is calculated as the final flux minus the initial flux. The negative sign indicates a decrease in flux over time.