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Xelga [282]
4 years ago
12

A circular conducting loop of radius 23.0 cm is located in a region of homogeneous magnetic field of magnitude 0.500 T pointing

perpendicular to the plane of the loop. The loop is connected in series with a resistor of 265 Ω. The magnetic field is now increased at a constant rate by a factor of 2.50 in 23.0s. Calculate the magnitude of the induced emf in the loop while the magnetic field is increasing.
Physics
1 answer:
Alchen [17]4 years ago
4 0

Answer:

The magnitude of the induced emf is

Ф = 5.419 x 10⁻³ V

Explanation:

Given:

r = 23.0 cm = 0.23 m

β₁ = 0.50 T

β₂ = 2.50 * 0.50 T = 1.25 T

Calculate the magnitude of the induced emf the magnetic field is increasing

Ф = A * Δβ / Δt

A = π * r² = π * 0.23² m = 0.166 m²

Ф = 0.166m² * (1.25 - 0.5) T / 23 s

Ф = 5.419 x 10⁻³ V

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Answer:

Therefore the required solution is

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Given vibrating system is

u''+\frac{1}{4}u'+2u= 2cos \omega t

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Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

Equating the coefficient of sinωt and cos ωt

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0.........(1)

and

\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0........(2)

Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

Therefore the required solution is

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make u the subject of the equation

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From the question,

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Substitute these values above into equation 3

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Hence, The magnification of the ornament is 0.25

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