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kolbaska11 [484]
3 years ago
12

A closed loop conductor that forms a circle with a radius of 2.0 m is located in a uniform but changing magnetic field. If the m

aximum emf Induced in the loop is 5.0 V what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Physics
1 answer:
Vladimir79 [104]3 years ago
5 0

Answer:

The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

Explanation:

Given;

radius of the circular loop, r = 2.0 m

maximum induced emf, E = 5.0 V

The emf induced in a magnetic field is given as;

emf = \frac{d\phi}{dt} \\\\\phi = AB\\\\emf = A\frac{dB}{dt} \\\\\frac{dB}{dt}  = \frac{emf}{A} \\\\where;\\A \ is \ the \ area \ circular \ l00p = \pi r^2 = \pi (2)^2 = 4\pi \ m^2\\\\\frac{dB}{dt}  = \frac{5}{4\pi} \\\\\frac{dB}{dt}  = 0.398 \ T/s

Therefore, the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi
alisha [4.7K]

Answer:

The total number of revolution is 50 rev.

Explanation:

Given that,

Angular speed = 5.0 rev/s

Time = 8.0 s

We need to calculate the angular acceleration

Using equation of angular motion

\omega_{f}-\omega_{i}=\alpha t

Put the value into the formula

5.0-0=\alpha\times8.0

\alpha=\dfrac{5.0}{8.0}

\alpha=0.625\ rev/s^2

We need to calculate the angular displacement

Using equation of angular motion

\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2

Put the value into the formula

\theta=0+\dfrac{1}{2}\times0.625\times(8.0)^2

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Now, The washer coming to rest from top spin

We need to calculate the angular acceleration

Using equation of angular motion

\omega_{f}-\omega_{i}=\alpha t

\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}

\alpha=\dfrac{0-5}{12}

\alpha=−0.4167\ rev/s^2

We need to calculate the angular displacement

Using formula of displacement

\theta'=\omega_{i}t+\dfrac{1}{2}\alpha t^2

Put the value into the formula

\theta'=5\times12+\dfrac{1}{2}\times(-0.4167)\times12^2

\theta'=30\ rev

We need to calculate the total number of revolution

\theta''=\theta+\theta'

\theta''=20+30

\theta''=50\ rev

Hence, The total number of revolution is 50 rev.

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3 years ago
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