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Paraphin [41]
4 years ago
15

If a single circular loop of wire carries a current of 62 A and produces a magnetic field at its center with a magnitude of 1.20

10-4 T, determine the radius of the loop.
Physics
1 answer:
s2008m [1.1K]4 years ago
3 0

Given Information:  

Current in loop = I = 62 A

Magnitude of magnetic field = B = 1.20x10⁻⁴ T

Required Information:  

Radius of the circular loop = r = ?  

Answer:  

Radius of the circular loop = 0.324 m

Explanation:  

In a circular loop of wire with radius r and carrying a current I  induces a magnetic field B which is given by

B = μ₀I/2r

Please note that for an infinitely straight long wire we use 2πr whereas for circular loop we use 2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space

Re-arranging the equation yields

r = μ₀I/2B

r = 4πx10⁻⁷*62/2*1.20x10⁻⁴

r = 0.324 m

Therefore, the radius of this circular loop is 0.324 m

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rewona [7]

Answer:

The new time period is  T_2 =  3.8 \  s

Explanation:

From the question we are told that

  The period of oscillation is  T =  5 \ s

   The  new  length is  l_2  =  0.76  \ m

Let assume the original length was l_1 = 1 m

Generally the time period is mathematically represented as

         T  =  2 \pi   \sqrt{ \frac{ I }{ mgh } }

Now  I is the moment of inertia of the stick which is mathematically represented as

           I  =  \frac{m * l^2 }{12 }

So

        T  =  2 \pi   \sqrt{ \frac{  m * l^2 }{12 *   mgh } }

Looking at the above equation we see that

        T  \ \ \  \alpha  \ \ \  l

=>    \frac{ T_2 }{T_1}  =  \frac{l_2}{l_1}

=>    \frac{ T_2}{5} =  \frac{0.76}{1}

=>     T_2 =  3.8 \  s

3 0
4 years ago
when an apple falls towards the earth,the earth moves up to meet the apple.is this true ?if yes,why is the earth's motion not no
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Answer:

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3 years ago
An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens
Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

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\frac{1}{v} -\frac{1}{u} =\frac{1}{f}

Putting the values

\frac{1}{v} +\frac{1}{15} =\frac{1}{14}

v = 210 mm .

B )

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C )

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21 mm is the answer .

D )

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\frac{210}{15} \times \frac{D}{f_e}

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

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= 170 ( in two significant figures )

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