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

Answer:-

$\blue{\bigstar}$ Magnitude of magnetic field inside Solenoid $\large\leadsto\boxed{\tt\red{0.0056972 \: T}}$

• 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,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\purple{B = \dfrac{\mu_0 NI}{L}}}}$

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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➪ $\sf B = \dfrac{(4 \pi \times 10^{-7}) \times 1200 \times 3.60}{0.95}$

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➪ $\sf B = \dfrac{(4 \times 3.14 \times 10^{-7}) \times 1200 \times 3.60}{0.95}$

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➪ $\sf B = 12.56 \times 10^{-7} \times 1200 \times 3.78$

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➪ $\sf B = 12.56 \times 10^{-7} \times 4536$

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➪ $\sf B = 56972.16 \times 10^{-7}$

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★ $\large{\bf\pink{B = 0.0056972 \: T}}$

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Therefore, the magnitude of the magnetic field inside the solenoid is 0.0056972 T .

Answer 2

Answer :-

• Magnetic field (B) $\leadsto$ $\large{\boxed{\tt{\red{0.0056972 \:T}}}}$

Given :-

• Length of solenoid ($\ell$) = 95cm = 0.95m

• Radius of solenoid (r) = 2cm = 0.02m

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$(not required in this question)

• No. of turns (N) = 1200

• Current ($\bold{I}$) = 3.60A

To find :-

• Magnitude of magnetic field (B)

Formula used :-

• $\boxed{\purple{\sf{B = \dfrac{\mu _{0} NI}{\ell}}}}$

Solution :-

$\sf{➪\:B = \dfrac{\mu _{0} NI}{\ell}}$

$\sf{➪\:B = \dfrac{(4\pi \times 10^{-7}) \times 1200 \times 3.60}{0.95}}$

$\sf{➪\:B = (12.56\times 10^{-7}) \times 1200 \times 3.78}$

$\sf{➪\:B = (12.56\times 10^{-7}) \times 4536}$

$\sf{➪\:B = 56972.16× 10^{-7}}$

$\sf{➪\:B =}$ $\large{\bold{\red{0.0056972 \:T}}}$