A solenoid is 95 cm long has a<br />radius of 2 cm and a winding of 1200 turns; it carries of 3.60 A, Calculate the magnitude of Magnetic field inside<br />Solenoid -
Answers
Answer:
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Step-by-step explanation:
Answer:-
★ Magnitude of magnetic field inside Solenoid {0.0057114}
0.0057114T
• 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,
★
B= L μ 0/ NI
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}B=
0.95
(4π×10
−7
)×1200×3.60
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➪ \sf B = \dfrac{(4 \times 3.14 \times 10^{-7}) \times 1200 \times 3.60}{0.95}B=
0.95
(4×3.14×10
−7
)×1200×3.60
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➪ \sf B = 12.56 \times 10^{-7} \times 1200 \times 3.78B=12.56×10
−7
×1200×3.78
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➪ \sf B = 12.56 \times 10^{-7} \times 4547.36B=12.56×10
−7
×4547.36
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➪ \sf B = 57114.94 \times 10^{-7}B=57114.94×10
−7
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★ {B = 0.0057114 \: T}}B=0.0057114T
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Therefore, the magnitude of the magnetic field inside the solenoid is 0.0057114 T .