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Musya8 [376]
3 years ago
11

A coil of 150 turns of wire is in the form of a circle of radius 7.5 cm, and sits in the center of a dipole magnet that produces

a uniform magnetic field of strength 1.5 T. The torque on the coil has a maximum value of 0.25 Nm. What is the current in the coils
Physics
1 answer:
irga5000 [103]3 years ago
7 0

Answer:

Current in the loop is 0.063 A        

Explanation:

Number of turns in the coil N = 150

Radius of the circular loop r = 7.5 cm = 0.075 m

So area A=\pi r^2=3.14\times 0.075^2=0.0176m^2

Magnetic field B = 1.5 T

Maximum torque is given \tau =0.25Nm

We have to find current in the coil

Torque on circular coil in magnetic field is equal to

\tau =BINA

0.25=1.5\times I\times 150\times 0.0176

I = 0.063 A

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Answer:

p(t)=\frac{1}{20}(1+2e^{-t/90})

Explanation:

Let p(t) be % of CO_2 at time t(time in minutes).

Amount of CO_2 in room=Volume of room \times p

Rate of change of CO_2 in room=Volume of room x \frac{dp}{dt}

=180\times \frac{dp}{dt}

Rate of inflow of CO_2=(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Inflow\ of \ air)

=(0.05\times \frac{1}{100})\times 2=\frac{1}{100}

Rate of CO_2 outflow

(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}

Therefore rate of change of CO_2 is \frac{1}{100}-\frac{p}{50}

% rate of change is100\times \ Rate of \ Change=\frac{1}{10}-2p

Therefore we have:

180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}

Integrate both:

\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}

\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C

In|20p-1|=-\frac{t}{90}+InK

+20In \ C=+In \ K

Raise both sides to base e,

20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})

Initially, there's only 0.25% of CO_2,

Now substitute p=0.25 and t=0

0.25=\frac{1}{20}(1+K)\\K=4

p=\frac{1}{20}(1+2e^{-t/90})

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A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 10 A, and that o
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Answer:

Part a)

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Part b)

P = 480 W

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Explanation:

Part a)

Power consumed by blow dryer is given as

P = iV

here supply voltage is given as

V = 120 volts

current rating of blow dryer is given as

i = 10 A

P = (120 V)(10 A)

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Part b)

Power consumed by vacuum cleaner is given as

P = iV

here supply voltage is given as

V = 120 volts

current rating of vacuum cleaner is given as

i = 4 A

P = (120 V)(4 A)

P = 480 W

Part c)

Energy consumed in a given interval of time is given as

Energy = power \times time

now we have to find the ratio of energy consumed in time interval of 11 minutes

so we have

\frac{E_{bd}}{E_{vc}} = \frac{P_{bd} t}{P_{vc} t}

now we have

\frac{E_{bd}}{E_{vc}} = \frac{1200 W\times 11 min}{480 W \times 11 min}

\frac{E_{bd}}{E_{vc}} = 5 : 2

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