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GuDViN [60]
3 years ago
9

6.7.1 Current and Current Density A sphere of radius 10 mm that carries a charge of 8 nC =8x 10°C is whirled in a circle at the

end of an insulated string. The rotation frequency is 100x rad/s. (a) What is the basic definition of current in terms of charge? (b) What average current does this rotating charge represent? (c) What is the average current density over the area traversed by the sphere?​
Physics
1 answer:
aev [14]3 years ago
6 0

Answer:

Part a)

Rate of charge flow is known as electric current

Part b)

Average current flow is

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

j = 1.27 \times 10^{-3} A/m^2

Explanation:

As we know that angular frequency of rotation is

\omega = 100\pi rad/s

now by basic definition of electric current

Part a)

Rate of charge flow is known as electric current

i = \frac{dq}{dt}

so here we have

i = \frac{Q}{T}

i = Qf

Part b)

here we know that

\omega = 2\pi f

100\pi = 2\pi f

f = 50 Hz

now we have

i = (8\times 10^{-9})(50)

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

So we have

j = \frac{i}{A}

j = \frac{4\times 10^{-7}}{\pi(10\times 10^{-3})^2}

j = 1.27 \times 10^{-3} A/m^2

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This question can be solved using the concept of friction energy.

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What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

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2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

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                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

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