Class 12 Electromagnetic Induction Mock Test (CBSE)

Topics in Electromagnetic Induction

• Experiments of Faraday and Henry
• Magnetic flux
• Faraday's law of induction
• Lenz's law and conservation of energy
• Motional electromotive force
• Inductance
• Mutual inductance
• Self-inductance
• AC generator

Electromagnetic Induction practice questions with answers

1. Q1Multiple choice·1 mark
   The frequency of alternating voltage generated by an AC generator can be increased by:

   1. (a)increasing the number of turns in the armature coil
   2. (b)increasing the strength of the magnetic field
   3. (c)decreasing the cross-sectional area of the coil
   4. (d)increasing the speed of rotation of the armature coil
   Show answer and step marks

   1. Correct option: (d) increasing the speed of rotation of the armature coil, as the frequency of the generated AC is directly proportional to the angular speed of rotation ($\nu = \omega / 2\pi$).+1

   Answer: (d)
2. Q2Multiple choice·1 mark
   In the foundational electromagnetic induction experiments conducted by Faraday and Henry, electric current is induced in a closed loop whenever:

   1. (a)there is relative motion between a bar magnet and the coil
   2. (b)the coil is kept completely stationary in a strong uniform magnetic field
   3. (c)a steady direct current flows through a nearby stationary coil
   4. (d)the magnetic flux linked with the coil remains constant over time
   Show answer and step marks

   1. Correct option: (a) there is relative motion between a bar magnet and the coil, leading to a time-varying magnetic flux through the coil which induces current.+1

   Answer: (a)
3. Q3Short answer·2 marks
   Define magnetic flux. A uniform magnetic field of magnitude $0.5\text{ T}$ makes an angle of $60^{\circ}$ with the normal to a rectangular surface of area $0.2\text{ m}^2$. Calculate the magnetic flux linked with the surface.
   Show answer and step marks

   1. Definition: magnetic flux through a surface is the product of the area and the component of $\vec{B}$ normal to it, $\Phi = \vec{B} \cdot \vec{A} = B A \cos\theta$ (in general $\Phi = \int \vec{B} \cdot d\vec{A}$); SI unit weber.+1
   2. Substituting the given values and calculating: $\Phi = 0.5 \times 0.2 \times \cos(60^{\circ}) = 0.1 \times 0.5 = 0.05\text{ Wb}$+1

   Answer: Magnetic flux $\Phi = \vec{B} \cdot \vec{A} = BA\cos\theta$ (the field component normal to the surface times its area); here $\Phi = 0.05\text{ Wb}$
4. Q4Short answer·3 marks
   A circular coil of radius $10\text{ cm}$, $500$ turns, and total resistance $2\ \Omega$ is placed with its plane normal to a uniform magnetic field of $0.4\text{ T}$. If the field is directed out of the page and drops to zero in $0.05\text{ s}$, calculate the magnitude of the induced emf and the total charge that flows through the coil during this time.
   Show answer and step marks

   1. Flux through each turn $\Phi = B A \cos\theta$ with $\theta = 0$ and $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi\text{ m}^2$; flux linkage $= N\Phi$.+1
   2. Average induced emf by Faraday's law: $|e| = \frac{N A |\Delta B|}{\Delta t} = \frac{500 \times 0.01\pi \times 0.4}{0.05} = 40\pi\text{ V} \approx 125.7\text{ V}$ ($125.6$ V with $\pi = 3.14$).+1
   3. Charge $Q = \frac{|e| \Delta t}{R} = \frac{N |\Delta\Phi|}{R} = \frac{40\pi \times 0.05}{2} = \pi\text{ C} \approx 3.14\text{ C}$.+1

   Answer: Average induced emf $= 40\pi\text{ V} \approx 125.7\text{ V}$ ($125.6$ V with $\pi = 3.14$); total charge $= \pi\text{ C} \approx 3.14\text{ C}$

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