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51.	The galvanometer shows deflection even when the bar magnet is stationary.(a) True(b) FalseI got this question by my school teacher while I was bunking the class.Origin of the question is Faraday and Henry Experiments topic in section Electromagnetic Induction of Physics – Class 12
Answer» RIGHT answer is (b) False To explain I WOULD say: No, this STATEMENT is false. When the bar magnet is STATIONARY, the pointer shows no deflection and the motion lasts only till the magnet is in motion. This shows that only the relative motion between the magnet and the coil is responsible for the generation of CURRENT in the coil.

52.	Calculate the mutual inductance between two coils if a current 10 A in the primary coil changes the flux by 500 Wb per turn in the secondary coil of 200 turns.(a) 10 H(b) 10^4 H(c) 1000 H(d) 100 HThe question was posed to me in homework.I'd like to ask this question from Electromagnetic Inductance topic in section Electromagnetic Induction of Physics – Class 12
Answer» Right option is (B) 10^4 H For EXPLANATION I would say: NΦ = MI 200 × 500 = M × 10 M = 10^4 H

53.	A metal rod is forced to move with constant velocity along two parallel metal rails, connected with a strip of metal at one end across a magnetic field (B) of 0.5 T, pointing out of the page. The rod is of length 45 cm and the speed of the rod is 70 cm/s. The rod has a resistance of 10 Ω and the resistance of the rails and connector is negligible. What is the rate at which energy is being transferred to thermal energy?(a) 0.225 W(b) 22.55 W(c) 2.25 × 10^-4 W(d) 2.25 × 10^-3 WThis question was addressed to me in homework.Origin of the question is Electromagnetic Induction in section Electromagnetic Induction of Physics – Class 12
Answer» Right choice is (d) 2.25 × 10^-3 W Easy explanation: Given: B = 0.5 T; v = 70 cm/s = 70 × 10^-2 m/s; L = 45 cm = 45 × 10^-2 m; R = 10 Ω Motional EMF (E) = vLB = 70 × 10^-2 × 45 × 10^-2 × 0.5 E = 0.15 V Current (I) = \(\frac {E}{R}\) I = \(\frac {0.15}{10}\) I = 0.015 A RATE of ENERGY or power (P) = I^2R P = 0.015^2 × 10 P = 2.25 × 10^-3 W Therefore, the rate of energy TRANSFER is 2.25 × 10^-3 W.

54.	Find the true statement.(a) The force on a current carrying conductor in a perpendicular magnetic field is due to the thermal energy dissipated(b) The force on a current carrying conductor in a perpendicular magnetic field is due to the mechanical energy dissipated(c) The force on a current carrying conductor in a perpendicular magnetic field is due to the current passing through it(d) The force on a current carrying conductor in a perpendicular magnetic field is due to the electrical energy initially converted from mechanical energyThe question was asked during a job interview.The question is from Energy Consideration: A Quantitative Study in division Electromagnetic Induction of Physics – Class 12
Answer» RIGHT option is (d) The FORCE on a current CARRYING conductor in a PERPENDICULAR magnetic FIELD is due to the electrical energy initially converted from mechanical energy To explain: The force on a current carrying conductor in a perpendicular magnetic field is due to drift velocity of charges along the current carrying conductor and the consequent Lorentz force acting on it. So, it is due to the electric energy which was the energy that mechanical energy dissipated in the system was initially converted to.

55.	A closed-loop move normal to the constant electric field between the plates of a large capacitor. What is the amount of current produced when it is wholly inside the region between the capacitor plates?(a) Maximum(b) Minimum(c) Zero(d) Independent of currentThe question was asked in exam.Question is from Electromagnetic Induction topic in portion Electromagnetic Induction of Physics – Class 12
Answer» Right CHOICE is (c) Zero Easiest explanation: No current is produced when it is wholly INSIDE the region between the capacitor plates. Current cannot be induced by CHANGING electric flux. So, the AMOUNT of current produced when it is wholly inside the region between the capacitor plates is zero.

56.	When is the magnetic flux said to be negative?(a) θ = 180^o(b) θ = 360^o(c) θ = 90^o(d) θ = 0^oThis question was addressed to me in an internship interview.This intriguing question originated from Electromagnetic Induction in portion Electromagnetic Induction of Physics – Class 12
Answer» Right OPTION is (a) θ = 180^o Explanation: If the normal points in the opposite direction of the field, then θ = 180^o and the FLUX is taken as negative. If the magnetic field is pointing opposite the direction of the SURFACE area then the value of this flux is negative.

57.	Which among the following affects the deflection in the galvanometer?(a) Area of the coil(b) Current passing through the coil(c) Speed with which the bar magnet is pulled towards or away from the coil(d) Resistance offered for current flowI got this question in an online quiz.My doubt stems from Faraday and Henry Experiments in chapter Electromagnetic Induction of Physics – Class 12
Answer» Correct choice is (c) Speed with which the bar MAGNET is pulled TOWARDS or away from the coil For explanation I would SAY: The deflection of the pointer is larger or smaller DEPENDING upon the speed with which the bar magnet is pulled towards or away from the coil. Moreover, the direction of deflection of the pointer depends upon the direction of motion of the bar magnet.

58.	Eddy currents may not be suitable for electrostatic shielding.(a) True(b) FalseThis question was addressed to me in examination.The origin of the question is Electromagnetic Induction topic in division Electromagnetic Induction of Physics – Class 12
Answer» Correct answer is (b) False Easy explanation: No, Eddy currents can be USED for electromagnetic SHIELDING. Electromagnetic shielding is the practice of reducing the electromagnetic FIELD in a space by blocking the field with barriers made of conductive or MAGNETIC materials. The higher the CONDUCTIVITY of the sheet used, the better the shielding of the transient magnetic field.

59.	What is measured by the eddy currents induced in energy meters?(a) Electric potential(b) Electric induction(c) Electric power(d) Electric energyThis question was posed to me during a job interview.My question is from Electromagnetic Induction topic in section Electromagnetic Induction of Physics – Class 12
Answer» Right choice is (d) Electric energy The explanation is: In energy meters used for measuring electric energy, the EDDY currents INDUCED in an aluminum disc are made USE of. THEREFORE, electric energy is MEASURED by the eddy currents induced in energy meters.

60.	Eddy currents can be used to heat localized tissues of the human body. What is this branch of science referred to as?(a) Inductology(b) Eddy dynamics(c) Inductothermy(d) Eddy mechanicsThis question was posed to me during an interview for a job.My enquiry is from Electromagnetic Induction topic in division Electromagnetic Induction of Physics – Class 12
Answer» The correct option is (C) Inductothermy The best explanation: EDDY currents can be USED to heat localized tissues of the human body. This branch is called inductothermy. Inductothermy is a method of electrotherapeutics in which certain parts of the body of the patient are heated by an alternating, predominantly high-frequency electromagnetic FIELD, which INDUCES eddy currents in body tissues.

61.	Pick out the expression for power in a rectangular conductor from the following.(a) P=\(\frac {B^2 l^2 v}{R}\)(b) P=\(\frac {B^2 lv^2}{R}\)(c) P=\(\frac {B^2 l^2 v^2}{R}\)(d) P=\(\frac {B^2 l^2 v^2}{R^2}\)I had been asked this question by my school principal while I was bunking the class.This intriguing question originated from Energy Consideration: A Quantitative Study topic in section Electromagnetic Induction of Physics – Class 12
Answer» The CORRECT choice is (c) P=\(\frac {B^2 l^2 v^2}{R}\) Easy explanation: In a rectangular CONDUCTOR, the emf (E) is given ➔ Blv So, current (I) = \(\frac {E}{R} = \frac {Blv}{R}\) POWER (P) = I^2R P=\(\frac {B^2 l^2 v^2}{R}\) The work DONE is mechanical and this mechanical energy is dissipated as Joule heat.