Question

41. The length of a solenoid is 0.1 m and its diameter is very small. A wire is wound over in two layers, The number of turns in the inner layer is 50 and that on the outer layer is 40. The strength of current flowing in two layers in the same direction is 3 ampere. The magnetic induction in the middle of the solenoid will be 1) 3.4 x 10-31 2) 3.4 x 10-gauss 3) 3.4 x 103T 4) 3.4 x 10 gauss​

Answer 1

Answer:

3..........................

Answer 2

The magnetic induction in the middle of the solenoid will be $\bold{3.4 \times 10^{-3} \ T}$

Given:

Length of a solenoid = 0.1 m

Number of turns = 50 (inner layer) + 40 (outer layer) = 90 turns

Current = 3 A

To find:

Magnetic induction = ?

Formula used:

Magnetic induction in the middle of the solenoid:

$\bold{B=\mu_0ni}$

Where,

$\bold{\mu_0}$ = Vacuum permeability = $\bold{4\pi\times10^{-7} \ H/m}$

n = Number of turns = $\bold{\frac{Number \ of \ Turns}{Unit \ Length}}$

i = Current flowing through wire

Solution:

$\bold{B=\mu_0ni}$

$\bold{B=4\pi\times10^{-7}\times\frac{90}{0.1}\times3}$

$\bold{\therefore B =3.4\times10^{-3}\ T}$