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Liono4ka [1.6K]

2 years ago

5

A flexible loop has a radius of 10.5 cm and it is in a magnetic field of B = 0.117 T. The loop is grasped at points A and B and

stretched until its area is zero. It takes 0.243 s to close the loop. What is the magnetic flux ΦB through the loop before it is streched? Tries 0/20 What is the magnetic flux through the loop after it is stretched? Tries 0/20 What is the magnitude of the average induced electromotive force ε in the loop during the stretching process?

1 answer:

MAXImum [283]2 years ago

3 0

Explanation:

Given that,

Radius = 10.5 cm

Magnetic field = 0.117 T

Time = 0.243 s

After stretched, area is zero

(I). We need to calculate the magnetic flux through the loop before stretched

Using formula of magnetic flux

$\phi=B\times A$

$\phi=B\times \pi r^2$

Where, B = magnetic field

r = radius

Put the value into the formula

$\phi=0.117\times3.14\times(10.5\times10^{-2})^2$

$\phi=4.05\times10^{-3}\ Tm^2$

(II). We need to calculate the magnetic flux through the loop after stretched

$\phi=B\times A$

Here, A = 0

$\phi=0$

So, The magnetic flux through the loop after stretched is zero.

(III). We need to calculate the magnitude of the average induced electromotive force

Using formula of the induced electromotive force

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

$\epsilon=-\dfrac{\phi_{after}-\phi_{before}}{t}$

$\epsilon=-\dfrac{0-4.05\times10^{-3}}{0.243}$

$\epsilon =16.67\times10^{-3}\ V$

Hence, This is the required solution.

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Where:

$m$ - Mass of the roller coaster car, in kilograms.

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$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

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