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CBSE 12th Standard Physics Subject Moving Charges And Magnetism 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

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Std: 12th Standard CBSE - Physics

chapter: - HOT Questions

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Questions:

Generic 2

5 x 2 = 10

A rigid circular loop of radius r and mass m lies in the x-y plane of a flat table and has a current I flowing in it. At this particular place the earth’s magnetic field is B = Bxi +Bzk. What is the value of I, so that loop starts tilting?

A straight wire of mass 200g and the length 1.5m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field B. What is the magnitude of B in tesla?

north to south direction.Specify the direction in which the uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

The vertical component of earth’s magnetic field at a given place is \(\sqrt { 3 } \) times its horizontal component. If the total intensity of earth’s magnetic field at a place is 0.4 G , find the value of horizontal component of earths field and angle of dip.

Two concentric circular coils A and B of radii 10 cm and 6 cm respectively, lie in the same vertical plane containing the north to south direction. coil A has 30 turns and carries a current of 10 A. Coil B has 40 turns and carries a current of 15 A .the sense of the current in A is anticlockwise and clockwise in B for an observer looking at the coils facing west. Give the magnitude and direction of net magnetic field.

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Answers Key:

Generic 2

5 x 2 = 10

M = IA = I r₂k
B = Bxi + BzK
T = M x B = (1\(\pi\) r_2k) x (Bxi + BzK)
= 1\(\pi\) R₂BXKxi = I r₂BXj
Torque due to the weight of the loop = mgr
I r²Bx = mgr Hence I = mg/\(\pi\)rB_x

For equilibrium of the wire in mid – air, weight of the wire = force exerted by magnetic field
Mg - I₁B sin 90⁰
B = mg/I₁ = \(\frac { 200\times { 10 }^{ -3 }\times 9.8 }{ 2\times 1.5 } =0.65T\)

Due to the electrostatic field electron will be deflected towards north. To keep it neutralized the magnetic force should deflect it towards south. For this purpose the magnetic field is to be applied perpendicular to the plane of the paper inward i.e vertically downward

Tan \(\eth \) = V/H = \(\sqrt { 3 } \)
\(\eth \) = 60
As V = \(\surd 3H\) and
B² = V² + H² = 3H² + H² = 4H²
(0.4)² = 4H² therefore
H = 0.2 G

Coil A - R₁ = 0.1m, n₁ = 30, I₁ = 10A
B1 = \({ \mu }_{ 0 }\) n₁I₁/2r₁ = 6 \(\pi \) x 10^-4 T directed towrds east
Coil B - R₂ = 0.6 m, n₂ = 40 , I₂ = 15 A
B₂ = \({ \mu }_{ 0 }\) n₂I₂/2r₂ = 2\(\pi \) x 10^-4T directed towards west
Net field B = B2 - B1 = (20\(\pi \) - 6\(\pi \)) 10^-4 T towards west