Question

A solenoid is 95 cm long has a
radius of 2 cm and a winding of 1200 turns; it carries of 3.60 A, Calculate the magnitude of Magnetic field inside
Solenoid -​

Answer 1

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Answer 2

Step-by-step explanation:

Answer:-

$[tex]\blue{\bigstar}★ Magnitude of magnetic field inside Solenoid \large\leadsto\boxed{\tt\red{0.0057114 \: T}}⇝ </p><p>0.0057114T</p><p> </p><p>$[/tex]

• Given:-

Length of a solenoid = 95 cm = 0.95 m

Radius of solenoid = 2 cm

Number of turns = 1200

Current = 3.60 A

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• To Find:-

Magnitude of magnetic field inside the solenoid.

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• Solution:-

We know,

$[tex]\pink{\bigstar}★ \large\underline{\boxed{\bf\purple{B = \dfrac{\mu_0 NI}{L}}}} </p><p>B= </p><p>L</p><p>μ </p><p>0$

NI[/tex]

where,

B = Magnetic field

N = Number of turns

I = Current

L = Length of solenoid

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• Substituting the values in Formula:-

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$[tex]➪ \sf B = \dfrac{(4 \pi \times 10^{-7}) \times 1200 \times 3.60}{0.95}B= </p><p>0.95</p><p>(4π×10 </p><p>−7</p><p> )×1200×3.60$

[/tex]

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$➪ \sf B = \dfrac{(4 \times 3.14 \times 10^{-7}) \times 1200 \times 3.60}{0.95}B= </p><p>0.95</p><p>(4×3.14×10 </p><p>−7</p><p> )×1200×3.60</p><p>$

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$➪ \sf B = 12.56 \times 10^{-7} \times 1200 \times 3.78B=12.56×10 </p><p>−7</p><p> ×1200×3.78$

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$[tex]➪ \sf B = 12.56 \times 10^{-7} \times 4547.36B=12.56×10 </p><p>−7</p><p> ×4547.36$

[/tex]

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$[tex]➪ \sf B = 57114.94 \times 10^{-7}B=57114.94×10 </p><p>−7</p><p>$

[/tex]

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$[tex]★ \large{\bf\pink{B = 0.0057114 \: T}}B=0.0057114T$[/tex]

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Therefore, the magnitude of the magnetic field inside the solenoid is 0.0057114 T .