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CBSE 12th Standard Physics Subject Moving Charges And Magnetism HOT Questions 1 Mark Questions 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

The hysteresis loss for a sample of 6 kg is 150 J/M² /cycle. If the density of iron is 7500 kg/m3 , calculate the energy loss per hour at 40cycle

Draw diagrams to show behavior of magnetic field lines near a bar of (i) Alluminium (ii) copper and (iii) mercury cooled to a very low temperature 4.2 K

When current is flowing through two parallel conductors in the same direction they attract while two beams of electrons moving in the same direction repel each other. Why?

Find the magnetic moment of a wire of length l carrying current I bent in the form of a circle

A copper wire is bent into a square of each side 6cm.If a current of 2A is passed through a wire what is the magnetic field at the centre of the square?

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Answers Key:

Generic 2

5 x 2 = 10

olume of sample = mass/density = 6/7500 m³
Energy loss/cycle = energy loss per volume /cycle x (volume ) = 150 x 6/7500
Energy loss/sec = 150 x 6 x 40 /7500
Energy loss/hour = 150 x 6 x 40 x 60/7500 J
= 1.728 x 10⁴ J

(i) Alluminium --- Paramagnetic

(ii) Copper and mercury = diamagnetic

Two conductors carrying current in same direction produce magnetic field and hence they attract. While two electron beams moving in the same directions repel due to its electric field (electrostatic force)

M = IA = I x \(\pi \) r²
But I = 2 \(\pi \) r, i.e r = 1/2 \(\pi \)
M = \(\pi \)² /4 \(\pi \)

B1 = 4 x \({ \mu }_{ 0 }\) I/4\(\pi\) a/2 (sin 45 + sin 45)
= 4 x \({ \mu }_{ 0 }\) 2/4 \(\pi\) x 3 (1 / 1.414 + 1/414 )
= 2\({ \mu }_{ 0 }\)/3\(\pi\) (1/.414 ) + 1/414 ) T