i have difficulty in understanding the different magnitudes of the magnetic field due to a current carrying solenoid; it is zero outside the solenoid but is strong along its axis; how can this be explained?

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
14 years ago

Dear Akshaya Karthikeyan,

Ans:- Look at the fig below. Neglect the field lines only consider the flow direction of the current. It is coming out of the plane of the paper in the upper portion and going into the plane of the paper in the lower portion.

Consider any  point inside the solinoid on it's axis. I hope you understand the right hand grip rule to determine the direction of the magnetic field for a given current direction. So see that inside the solinoid, the field is towards right i.e towards the tip of the arrows. But consider a point at the bottom outside the solinoid. Here if you apply the Grip rule then, You can see that it is towards right for the upper portion but is towards the left for the lower portion . TRY IT.

So if the diameter of tae solinoid is very thin then these fields are equal and opposite hence cancel out each other.Inside the solinoid it's value can be calculated using Ampere's circuital law. Let the value be B. Then we get

∫B•dl = µNI

where N =total no of turns

As we have seen that, the value of The magnetic field B is along the length Hence B•dl=B dl

So ∫B•dl=B L

where L= total length

BL=µNI

B=µNI/L

B=µni

where n= no of turns per unit length

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