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CBSE 12th Standard Physics Subject Electromagnetic Induction HOT Questions 1 Mark Questions With Solution 2021

Published on: 27/05/2021

QB365 Provides the HOT Question Papers for Class 12 Physics, and also provide the detail solution for each and every HOT Questions. HOT Questions will help to get more idea about question pattern in every exams and also will help to get more marks in Exams

Download CBSE Class 12th Standard CBSE Physics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 12th Standard CBSE Physics

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Std: 12th Standard CBSE - Physics

chapter: - HOT Questions

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Questions:

Generic 2

5 x 2 = 10

A copper ring is held horizontally and bar magnet is dropped through the ring with its length along the axis of the ring. Will the acceleration of the falling magnet be equal to, greater than or less than that due to gravity?

A magnet is moved in the direction indicated by an arrow between two coils A and B as shown below. Suggest the direction of induced current in each coil L.

A bar of mass M is suspended by two springs as shown below. Assume that a magnetic field B is directed out of the page. Each spring has a spring constant K. Describe the bar’s displacement when a current I is sent through it in the direction shown

A bar PQ of mass M is suspended by two wires as shown below. Assume that a uniform magnetic field B is directed into the page. Find the tension in each supporting wire when the current through the bar is I.

A particle with charge ‘q’ and mass ‘m’ is shot with kinetic energy K into the region between two plates as shown in the figure. If the magnetic field between the plates is B and as shown, how large must B be if the particle is to miss collision with the opposite plate?

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Answers Key:

Generic 2

5 x 2 = 10

Less than that due to gravity.

Due to Lenz’s law, end A will behave as South Pole and end B will behave as North Pole. The end face A will have clock wise direction of current and end face B will have anti clock wise direction of current when seen from the magnet side

Due to Flemings Right hand Rule the magnetic force ILB is directed downward. This constant force shifts the equilibrium position downward by a displacement = (ILB)/2K

According to the Flemings Right-hand Rule, the magnetic force ILB is directed upward. Equilibrium in the vertical direction yields 2T + ILB = Mg, so that T = (Mg – ILB)/²

Just to miss the opposite plate, the particle must move in a circular path with radius d so that Bqv = mv2/d, B = (2mK)1/2/(qd)