Physics, asked by akash8647, 8 months ago

A solenoid 15cm long has 300 truns of wire . A current of 5A flows through. What is magnitude of magnetic field inside solenoid
A 1.3x10 wb
B. 13×10 wb
C none​

Answers

Answered by Anonymous
38

Given :

▪ Length of solenoid = 15cm

▪ No. of turns = 300

▪ Current flow = 5A

To Find :

▪ Magnetic field inside the solenoid.

CalculaTioN :

✴ Magnetic field due to straight solenoid, at a point inside the solenoid is given by

\bigstar\underline{\boxed{\bf{\red{B=\mu_o nI}}}}

Here n is the number of turns per unit length.

\dashrightarrow\sf\:B=\mu_o nI\\ \\ \dashrightarrow\sf\:B=(4\pi\times 10^{-7})\times \dfrac{300}{0.15}\times 5\\ \\ \dashrightarrow\sf\:B=12.56\times 10^{-7}\times 10^{4}\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{B=1.3\times 10^{-2}\:T}}}}\:\orange{\bigstar}

Answered by Anonymous
9

Given that ,

  • Length of solenoid (l) = 15 cm
  • Number of turns (N) = 300
  • Current (I) = 5 Amp

We know that , the magnetic field inside a solenoid is given by

 \sf \large  \fbox{B =  \mu_{o} nI}

Where ,

B = magnetic field

N = No. of turns per unit length

I = Current

Thus ,

 \sf \mapsto B = 4\pi \times  {(10)}^{ - 7}  \times  \frac{300}{15 \times  {(10)}^{ - 2} }  \times 5 \\  \\  \sf \mapsto B = 4 \times 3.14 \times  {(10)}^{ - 3}  \times  \frac{3}{3}  \\  \\  \sf \mapsto B = 12.56 \times  {(10)}^{ - 3} \:  \:  testla

 \therefore \sf \underline{The  \: magnetic  \: field \:  inside \:  a \:  solenoid  \: is  \: 12. 56  \times  {(10)}^{ - 3}  \:  t}

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