Question

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​

Answer 1

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.

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

Answer 2

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 ,

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

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