Magnetic Fields and Electromagnetic Induction - Capacitance and Dielectrics MCQ & Objective Questions
The topics of Magnetic Fields and Electromagnetic Induction, along with Capacitance and Dielectrics, are crucial for students preparing for various school and competitive exams in India. Understanding these concepts not only enhances your theoretical knowledge but also improves your problem-solving skills. Practicing MCQs and objective questions on these topics is an effective way to boost your exam preparation and score better in important exams.
What You Will Practise Here
Fundamental concepts of magnetic fields and their properties.
The principles of electromagnetic induction and Faraday's law.
Capacitance: definitions, formulas, and applications in circuits.
Understanding dielectrics and their role in capacitors.
Key diagrams illustrating magnetic field lines and capacitor setups.
Problem-solving techniques for numerical questions related to these topics.
Important questions and previous years' exam patterns for better preparation.
Exam Relevance
The topics of Magnetic Fields and Electromagnetic Induction, as well as Capacitance and Dielectrics, frequently appear in CBSE, State Boards, NEET, and JEE exams. Students can expect questions that test both conceptual understanding and numerical problem-solving skills. Common question patterns include direct application of formulas, conceptual MCQs, and scenario-based problems that require critical thinking.
Common Mistakes Students Make
Confusing the direction of magnetic fields and electric fields.
Misapplying formulas related to capacitance in different configurations.
Overlooking the significance of dielectric materials in capacitors.
Failing to interpret graphical representations of magnetic fields correctly.
Neglecting units and conversions in numerical problems, leading to incorrect answers.
FAQs
Question: What is the formula for calculating capacitance? Answer: The capacitance (C) is calculated using the formula C = Q/V, where Q is the charge stored and V is the voltage across the capacitor.
Question: How does electromagnetic induction work? Answer: Electromagnetic induction occurs when a changing magnetic field induces an electromotive force (EMF) in a conductor, as described by Faraday's law.
Ready to enhance your understanding? Dive into our practice MCQs and test your knowledge on Magnetic Fields and Electromagnetic Induction - Capacitance and Dielectrics. Your success in exams starts with consistent practice!
Q. According to Coulomb's law, what is the force between two charges of +3 µC and -2 µC separated by a distance of 0.1 m?
A.
0.54 N
B.
0.60 N
C.
0.72 N
D.
0.80 N
Solution
Force F = k * |q₁ * q₂| / r² = (8.99 x 10^9 N m²/C² * |3 x 10^-6 C * -2 x 10^-6 C|) / (0.1 m)² = 0.54 N.
Q. If the distance between two point charges is doubled, how does the force between them change according to Coulomb's law?
A.
It doubles
B.
It quadruples
C.
It halves
D.
It becomes one-fourth
Solution
According to Coulomb's law, the force F between two point charges is inversely proportional to the square of the distance r between them. If the distance is doubled, the force becomes F/4.
Q. What is the capacitance of a parallel plate capacitor with an area of 2 m² and a separation of 0.01 m filled with a dielectric of relative permittivity 5?
A.
0.88 µF
B.
1.77 µF
C.
2.65 µF
D.
3.54 µF
Solution
Capacitance C = (ε₀ * ε_r * A) / d = (8.85 x 10^-12 F/m * 5 * 2 m²) / 0.01 m = 8.85 x 10^-11 F = 1.77 µF.
Q. What is the dielectric constant of a material if a capacitor with air as dielectric has a capacitance of 10 µF and the same capacitor with the material has a capacitance of 30 µF?
