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Physics NCERT Exemplar Solutions Class 11th Chapter Ten Electromagnetic Induction Physics Ncert Solutions Class 12th Physics Class 12th

A small bar magnet is moved through a coil at constant speed from one end to the other. Which of the following series of observations will be seen on the galvanometer G attached across the coil?

Three positions shown describe:

(a) the magnet's entry

(b) magnet is completely inside and

(c) magnet's exit

Option 1 -

Option 2 -

Option 3 -

Option 4 -

1 View | Posted a month ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

(A) The magnet's entry

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An aeroplane, with its wings spread 10m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth’s field at that part is 2.5 × 10^-⁴Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be…………..

alok kumar singh

B_v = B sin 60°

-> $B_{v} = 2.5 \times 1 0^{- 4} \times \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} \times v \times l = 2.5 \times 1 0^{- 4} \times \frac{\sqrt[]{3}}{2} \times 180 \times \frac{5}{18} \times 1 = 108.25 \times 1 0^{- 3} v o l t s$

Two conducting circular loops of radii R₁ and R₂ are placed in the same plane with their centres coinciding. If R₁ » R₂, the mutual inductance M between them will be directly proportional to:

alok kumar singh

M = φ? /I? = (B? A? )/I? = [ (μ? I? /2R? )πR? ²]/I?
[Diagram of two concentric coils]
M = (μ? πR? ²)/ (2R? )
M ∝ R? ²/R?

Figure shows a circuit that contains four identical resistors with resistance R = $2.0 Ω,$ two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current ‘i’ just after switch ‘S’ is closed will be:

Vishal Baghel

R = $2 Ω$

L = 2 mH

E = 9V

$i = \frac{ε}{2 R} = \frac{9 v}{4 Ω} = 2.25 A$

Just after the switch ‘S’ is closed, the inductor acts as open circuit.

An inductor coil stores 64J of magnetic field energy nad dissipates energy at the rate of 640W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds:

Vishal Baghel

$U_{B} = \frac{1}{2} L l^{2} . . . . . . . . . . . . (i)$

P_R = l²R

=> 640 = (8)² R

R = $10 Ω$

from (i) 64 = $\frac{1}{2} \times L \times {(8)}^{2}$

L = 2H

$l = \frac{L}{R} = \frac{2}{10} = 0.2 s e c o n d$

A planar loop of wire rotates in a uniform magnetic field. Initially, at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

alok kumar singh

At any time $t$

$? = B A c o s ? \frac{π t}{5} (? ω = \frac{2 π}{T} = \frac{2 π}{10} = \frac{π}{5})$

$\frac{d ?}{d t} = - \frac{π}{5} B A s i n ? (\frac{π t}{5})$

$e = \frac{π}{5} B A s i n ? (\frac{π t}{5})$

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$t = 2.5 s e c$
e will be minimum when $\frac{π t}{5}$ is $π, 0$

$t = 0,5 s e c$

So, answer will be (1).

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