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horsena [70]
4 years ago
10

A circular loop has radius R and carries current I2 in a clockwise direction. The center of the loop is a distance D above a lon

g, straight wire.
What is the magnitude of the current I1 in the wire if the magnetic field at the center loop is zero? Express your answer in terms of the variables I2, R, D, and appropriate constants (μ0 and π).
Physics
1 answer:
romanna [79]4 years ago
5 0

Answer:

I_{1} = (πDI_{2})/R

Explanation:

If we define the magnitude of the field as B, then we have:

Total magnitude of the field B_{t} = magnitude of the field B_loop + magnitude of the field B_wire. The total magnitude is equivalent to zero. Therefore, the field B_loop has an inward direction while the field B_wire has an outward direction.

B_loop = (μ0)*(I_{2})/2*R

B_wire = (μ0)*(I_{1})/2*π*D

Thus:

B_loop = B_wire at the center of the loop.

(μ0)*(I_{2})/2*R = (μ0)*(I_{1})/2*π*D

I_{1} = (πDI_{2})/R

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Answer:

Power required will be 7.2 watt

Explanation:

We have given battery of Barble jeep is 6 volt

So potential difference V = 6 volt

Resistance of electric motor R = 5 ohm '

We have to find the power motor using to drive the jeep

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sergij07 [2.7K]

Hello!

The answer should be the third option.

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If an object is accelerating, it is traveling the same distance for each time interval of its motion
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Answer:

false

Explanation:

if an object is accelerating the object will not travel the same distance every time interval

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3 years ago
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8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-
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Answer: 0.333 h

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:  

A=A_{o}.2^{\frac{-t}{H}} (1)  

Where:  

A=0.375 g is the final amount of the material  

A_{o}=3 g is the initial amount of the material  

t=1 h is the time elapsed  

H is the half life of the material (the quantity we are asked to find)  

Knowing this, let's substitute the values and find h from (1):

0.375 g=(3 g)2^{\frac{-1h}{H}} (2)  

\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}}) (4)  

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Clearing H:

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