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ludmilkaskok [199]

3 years ago

11

A conducting loop of radius 1.50 cm and resistance 8 × 10−6Ω is perpendicular to a uniform magnetic field of magnitude 23.0 × 10

−6T. The field magnitude drops to zero in 7 ms. How much thermal energy is produced in the loop by the change in the magnetic field?

1 answer:

Anika [276]3 years ago

4 0

To solve this problem it is necessary to apply the concepts related to electromotive force or induced voltage.

By definition we know that the induced emf in the loop is equal to the negative of the change in the magnetic field, that is,

$\epsilon = -A \times \frac{\Delta B}{\Delta t}$

$\epsilon = -A \times (\frac{B_f-B_i}{t_f-t_i})$

Where A is the area of the loop, B the magnetic field and t the time.

Replacing with our values we have that

$\epsilon = -(\pi (1.5*10^{-2})^2)(\frac{0-23*10^{-6}}{7*10^{-3}-0})$

$\epsilon = 2.3225*10^{-6}V$

Therefore the thermal energy produced is given by

$E = P*t = \frac{\epsilon^2}{R}t$

$E = \frac{(2.3225*10^{-6})^2}{8*10^{-6}}*(7*10^{-3})$

$E = 4.719*10^{-9}J$

The thermal energy produced in the loop is $4.719*10^{-9}J$

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Answer

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You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

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Answer:

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$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

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