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5 months ago

Electromagnetic Induction Class 12th

In the given figure the magnetic flux through the loop increases according to the relation φB(t) = 10t² + 20t, where φB is in milliwebers and t is in seconds. The magnitude of current through R = 2Ω resistor at t = 5s is _____ mA.

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Vishal Baghel

Contributor-Level 10

E = -dφ/dt = - (20t + 20) mV.
At t=5s, E = - (100+20) = -120mV.
I = |E|/R = 120mV/2Ω = 60mA.

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New answer posted

5 months ago

Electromagnetic Induction Class 12th

A circular conducting coil of radius 1 m being heated by the change of magnetic field B passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2μΩ. The magnetic field is slowly switched off such that its magnitude changes in time as B = (4/π) * 10?³ T [1 - ( t/100 )]. The energy dissipated by the coil before the magnetic field is switched off completely is E = _______ mJ.

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Raj Pandey

Contributor-Level 9

ε = |dΦ/dt| = A|dB/dt|
Given B (t), dB/dt = (4/π) * 10? ³ * (-1/100)
ε = (π * 1²) * | (4/π) * 10? ³ * (-1/100)|
= 4 * 10? V
To find when B=0:
B = 0 ⇒ 1 - t/100 = 0
⇒ t = 100 second
Energy Dissipated, E = P * t = (ε²/R) * t
E = (4*10? )² / (2*10? ) * 100 = 80mJ

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New answer posted

5 months ago

Electromagnetic Induction Class 12th

An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 µs is x/100 mA. Then x is equal to _______. (Take e?¹ = 0.37)

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Raj Pandey

Contributor-Level 9

I = I? e? /?
= (20/10000) e^- (1*10? / 10*10? ³)
= 2 * 10? ³ e? ¹
The provided solution calculates as:
= 2 * 10? ³ e? ¹
= 2e? ¹ mA
= 2 * 0.37mA
= 0.74 mA = 74/100 mA
(Note: There seems to be a calculation discrepancy in the source image, the steps shown lead to I = 2e? ¹ mA)

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New answer posted

6 months ago

Electromagnetic Induction Class 12th

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………….

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Vishal Baghel

Contributor-Level 10

Bv = B sin 60°

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

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

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New answer posted

6 months ago

Electromagnetic Induction Class 12th

A coil of inductance 2H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, the energy stored in the coil after 4s is…………… J.

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Vishal Baghel

Contributor-Level 10

$L \frac{d i}{d t} = V \Rightarrow 2 d i = 3 t d t \Rightarrow 2 \int_{0}^{i} d i = 3 \int_{0}^{4} t d t \Rightarrow i = 12 A$
$U = \frac{1}{2} L i^{2} = \frac{1}{2} * 2 * 1 2^{2} = 144 J$

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New answer posted

6 months ago

Electromagnetic Induction Class 12th

A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:

A. hold the sheet there if it is magnetic.

B. hold the sheet there if it is non-magnetic.

C. move the sheet away from the pole with uniform velocity if it is conducting.

D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.

Choose the correct statement(s) from the options given below:

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alok kumar singh

Contributor-Level 10

A magnetic pole will repel or attract magnetic sheet so force is need.

B. If sheet is non-magnetic, no force needed.

C. If it is conducting, then there will be addy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed.

D. The non-conducting and non-polar sheet do not interact with magnetic field of magnet.

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New answer posted

6 months ago

Electromagnetic Induction Class 12th

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A . The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 π * 1 0^{- 7} SI$ units):

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alok kumar singh

Contributor-Level 10

The magnitude of magnetic field due to circular coil of $N$ turns is given by

$B_{C} = \frac{μ_{0} i N}{2 R}$

$= \frac{4 π * 1 0^{- 7} * 7 * 100}{2 * 0.1}$

$= 4.4 * 1 0^{- 3} T$

$= 4.4 mT$

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New answer posted

6 months ago

Electromagnetic Induction Class 12th

In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:

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alok kumar singh

Contributor-Level 10

North of magnet is moving away from solenoid 1 so end $B$ of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end $C$ of solenoid 2 is South.

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New answer posted

6 months ago

Electromagnetic Induction Class 12th

The arm PQ of a rectangular conductor is moving from x = 0 to x = 2b outwards and then inwards from x = 2b to x = 0 as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x = 0 to x = b. Identify the graph the variation of different quantities with distance.

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alok kumar singh

Contributor-Level 10

When arm PQ of a rectangular conductor is moving from x = 0 to x = b, the flux $(? = a x)$ linked with loop increases and while moving from x = b to x = 2b and from x = 2b to x = b, $(? = a b)$ flux remains constant and then flux $(? = a x)$ decreases to zero as it moves from x = b to x = 0.

$e_{i n} = - \frac{d ?}{d t} = - a \frac{d x}{d t} = - a v \Rightarrow$ Induced emf

$P = \frac{{(e_{i n})}^{2}}{R} \Rightarrow$ Power dissipated

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New answer posted

6 months ago

Electromagnetic Induction Class 12th

An object of mass 'm' is being moved with a constant velocity under the actin of an applied force of 2N along a frictionless surface with following surface profile.

The correct applied force vs distance graph will be:

0 Follower 11 Views

alok kumar singh

Contributor-Level 10

Object is moving in upward direction with constant velocity so in upward motion (+2N) and for downward motion (-2N) So option (1) is correct representation.

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