Electromagnetic Induction

A simple LR circuit is connected to a battery at t = 0. The time instant at which rate of energy storage in inductor is half of power delivered by battery

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Contributor-Level 10

i = (V? /R) (1 - e? /? )
U = ½Li²
dU/dt = Li (di/dt)
(R/2L) * (V? ²/R²) (1 - e? /? )²e? /? = (V? ²/R) (1 - e? /? )
t = (L/R)ln2

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a month ago

Metallic disc of inner radius R and outer radius 2R is rotating with an angular speed ω. What will be the electric potential difference that appears between the inner and outer periphery. The mass of electrons is m and charge is e.

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Contributor-Level 10

eE = e (dV/dr) = mω²r
ΔV = ∫dV = ∫ (m/e)ω²rdr from R to 2R
ΔV = (3mω²R²)/ (2e)

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a month ago

The magnetic energy stored in an inductor of inductance $4 μ H$ carrying a current of 2A is:

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Contributor-Level 9

Magnetic energy stored in an inductor $= \frac{1}{2} {L I}^{2}$

$\begin{array}{r} = \frac{1}{2} * 4 * 10^{- 6} * (2)^{2} \\ = 8 μ J \end{array}$

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a month ago

The area of rectangular field (in $m^{2}$ ) of length $55.3 m$ and breadth 25 m after rounding off the value of correct significant digits is:

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Contributor-Level 9

Area $=$ Length $*$ Breadth

$= 55.3 * 25$

$= 1382.5$

$= 14 * 10^{2}$

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a month ago

The dimensions $[{M L T}^{- 2} A^{- 2}]$ $[^{}^{}]$ belong to the:

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Contributor-Level 9

$[{M L T}^{- 2} A^{- 2}] =$ Magnetic permeability

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a month ago

The magnetic energy stored in an inductor of inductance $4 μ H$ carrying a current of $2 A$ is:

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alok kumar singh

Contributor-Level 10

For transformer

$(\frac{i_{p}}{i_{s}}) = (\frac{V_{s}}{V_{P}}) = (\frac{N_{s}}{N_{P}})$

For bulb, $V s = 12 V$

$\begin{array}{r} i_{s} = (\frac{60}{12}) = 5 A \\ \Rightarrow \frac{i_{p}}{5} = (\frac{12}{220}) \\ i_{P} = \frac{60}{220} = (\frac{3}{11}) = 0.27 A \end{array}$

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a month ago

A square loop of side $1 m$ and resistance $1 Ω$ is placed in a magnetic field of $0.5 T$ . If the plane of loop is perpendicular to the direction of a magnetic field, the magnetic flux through the loop is

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Contributor-Level 9

$B = 0.5 T$

Angle between $\overset{?}{B}$ & A $\overset{?}{A}$ is zero

$? =$ B.A. $c o s ? 0$

$= 0.5 * (1) * 1$

$= 0.5 W b$

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New answer posted

a month ago

The time taken for the magnetic energy to reach 25% of its maximum value, when solenoid of resistance R, inductance L is connected to a battery, is:

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Contributor-Level 10

I = (V/R) (1 - e^ (-t/τ) where τ = L/R

E = (1/2)LI² = (1/2)L (V²/R²) (1 - e^ (-t/τ)² ⇒ Energy stored in inductor

According to Question, we can write
(1/4) * (1/2) (LV²/R²) = (1/2) (LV²/R²) (1 - e^ (-t/τ)²
⇒ 1/4 = (1 - e^ (-t/τ)² ⇒ 1/2 = 1 - e^ (-t/τ) ⇒ e^ (-t/τ) = 1/2

⇒ t/τ = ln (2) ⇒ t = τln (2) = (L/R)ln (2)

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New answer posted

2 months ago

Two concentric circular coils, C₁ and C₂ are placed in the XY plane. C₁ has 500 turns, and a radius of 1 cm. C₂ has 200 turns and radius of 20 cm. C₂ carries a time dependent current I(t) = (5t² - 2t + 3)A where t is in s. The emf induced in C₁ (in mV ), at the instant t = 1 s is x/4. The value of x is

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alok kumar singh

Contributor-Level 10

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New answer posted

2 months ago

A series L – R circuit is connected to a battery of emf V. If the circuit is switched on at t = o, then the time at which the energy stored in the inductor reaches (1/n) times of its maximum value, is :

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