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2 years ago

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A loop of wire is placed in a uniform magnetic field. For what orientation of the loop is the magnetic flux a maximum? For what

orientation is the flux zero? Draw diagrams of these two situations. (Do this on paper. Your instructor may ask you to turn in these diagrams.)

1 answer:

andreyandreev [35.5K]2 years ago

8 0

Answer:

The flux is calculated as φ=BAcosθ. The flux is thereforemaximum when the magnetic field vector is perpendicular to theplane of the loop. We may also deduce that the flux is zero whenthere is no component of the magnetic field that is perpendicularto the loop.

when angle is zero then flux is maximium because when angle zerocos is maximium

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2 years ago

Find the ratio of the lengths of the two mathematical pendulums, if the ratio of periods is 1.5

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Answer:

The ratio of lengths of the two mathematical pendulums is 9:4.

Explanation:

It is given that,

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Let the lengths be L₁ and L₂.

The time period of a simple pendulum is given by :

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or

$T^2=4\pi^2\dfrac{l}{g}\\\\l=\dfrac{T^2g}{4\pi^2}$

Where

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or

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ATQ,

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1 year ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

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Answer:

$KE_A=33\ J$

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Explanation:

Given:

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Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

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$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

<u>Using the conservation of linear momentum:</u>

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

<u>Now the kinetic energy of particle B:</u>

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

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Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

<u>Now the kinetic energy of particle A:</u>

<u /> $KE_A=\frac{1}{2}m_A.v_A^2$ <u />

<u /> $KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$ <u />

$KE_A=33\ J$

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2 years ago

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Answer:

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