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2 months ago

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A square loop of wire, with sides of length a, lies in the first quadrant of the xy plane, with one corner at the origin. In thi

s region, there is a non uniform time-dependent magnetic field B(y, t) = ky3t2 ˆz (where k is a constant). Find the emf induced in the loop.

1 answer:

Softa [3K]2 months ago

5 0

Answer:

$emf=-\dfrac{1}{2}kta^5$

Explanation:

We have

B(y, t) = k y ³t²

To determine the total flux through the loop, we need to integrate over it

$\phi =\int B.dS$

Since the loop is square,

$\phi =\int B.dS$

B(y, t) = k y ³t²

$\phi =kt^2\int_{0}^{a}dx\int_{0}^{a}y^3dy$

$\phi =\dfrac{1}{4}kt^2a^5$

We know that the emf can be expressed as

$emf=-\dfrac{d\phi }{dt}$

$\phi =\dfrac{1}{4}kt^2a^5$

Thus

$emf=-\dfrac{1}{2}kta^5$

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Ostrovityanka [3204]

Answer:

The period of the pendulum measuring 16 m is double that of the 4 m pendulum.

Explanation:

Recall that the period (T) of a pendulum with length (L) is defined by:

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Since both pendulums are positioned at the same location, the value of "g" will be consistent for both, and when we compare the periods, we find:

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2 months ago

A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o

Maru [3345]

Answer:

The excess charge is $Q = 3.5 *10^{-7} \ C$

Explanation:

According to the question, we are informed that

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The radius of the sphere is

$r = \frac{d}{2}$

by plugging in given values

$r = \frac{0.45}{2}$

$r = 0.225 \ m$

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$V = \frac{k * Q }{r }$

Where k is Coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

Based on the question stating there are no other charges, Q represents the excess charge

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$Q = \frac{V* r}{ k}$

inserting the numerical values

$Q = \frac{14 *10^{3} 0.225}{ 9*10^9}$

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1 month ago

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Answer:

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b

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