Question

A 1 m long solenoid containing 1000 turns produces a magnetic flux density of 3.14 103 t inside it. The current flowing in the solenoid will be

Answer 1

Given:

Length of the solenoid = 1m

Number of turns = 1000

Magnetic flux density = 3.14103 T

To Find:

Using above parameters we have to find the value of current flowing inside the solenoid.

Solution:

We have given the magnetic flux density which is magnetic field strength (B).

To solve this solution first we have to find out the formula. Here we will use formula ,

$\bold{B = \mu_{0}nI}$     ….(1)

Here, B = magnetic flux density

n = number of turns per unit length

and I = current flowing in the solenoid

∴$n=\frac{1000}{1}\hspace{1mm} m^{-1} = 1000\hspace{1mm}m^{-1}$

∴Putting all the given values in equation (1)

$\bold{3.14103T = 4\pi \times 10^{-7}TmA^{-1}\times 1000m^{-1}\times I}$

$=> I =\frac{3.14103}{4\pi \times 10^{-4}} A$

$=>I=0.25\times10^{4}A=2500 A$

∴Current flowing in the solenoid will be 2500A.