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Jobisdone [24]
2 years ago
11

A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magni

tude of the magnetic field inside the solenoid.
Physics
1 answer:
irinina [24]2 years ago
7 0

The magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

The given parameters;

  • <em>length of the solenoid, L = 91 cm = 0.91 m</em>
  • <em>radius of the solenoid, r = 1.5 cm = 0.015 m</em>
  • <em>number of turns of the solenoid, N = 1300 </em>
  • <em>current in the solenoid, I = 3.6 A</em>

The magnitude of the magnetic field inside the solenoid is calculated as;

B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\

where;

\mu_o is the permeability of frees space = 4π x 10⁻⁷ T.m/A

B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T

Thus, the magnitude of the magnetic field inside the solenoid is 6.46 \times 10^{-3} \ T.

Learn more here:brainly.com/question/17137684

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John wants a new video tape and a new shirt, but he only has $12.00. He buys the video tape and gets $0.52 change. What was his
mestny [16]

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You have the answer in your comments. I will be copying it so your question doesn't get deleted.

The answers is $0.58

$11.48

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An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is
Lady_Fox [76]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

\\

\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

\\

<h3>☯ <u>Now, Finding the magnification </u></h3>

\\

\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

\\

\dashrightarrow\:\: \sf{m = -2}

\\

<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}

3 0
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The figure shows two springs (k1 = 10 N/m and k2 = 20 N/m ) attached to a block that can slide on a frictionless surface. In the
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Answer:

Explanation:

a )

Energy stored by left spring when compressed = 1/2 k x²

= .5 x 10 x .02² = .002 J .

Let compression in right spring = y

energy stored to right spring = 1/2 k y²

1/2 k y² = 0.002

.5 x 20 x y² = 0.002

y = .01414 m

= 1.4  cm

7 0
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