Question

What current is required in the windings of a long solenoid that has 420 turns uniformly distributed over a length of 0.575 m in order to produce inside the solenoid a magnetic field of magnitude 4.22×10−5 T? The permeablity of free space is 1.25664 × 10 Tm/A.

Answer 1

The current in the solenoid=0.463 Ampere

Step-by-step explanation:

Given:

Number of turns , N=420

Length of the solenoid=0.575 m

Magnitude of Magnetic Field,
`B=4.22*10^(-5)\ \rm T`

We know that the magnitude of the magnetic Field inside the solenoid is

`B=\mu_0 ni`

Where

• 
  `\mu_0` is the permeability of free space.
• n is the number of turns per unit length
• i is the current

According to question

`B=\mu_0 ni\\\\4.22*10^(-5)=1.24664*10^(-7)* (420)/(0.575)* i\\=0.463\ \rm Amp`

Hence the current is calculated.