2 consider a 1000 turn solenoid with a radius of 10 cm and a...

2. Consider a 1000-turn solenoid with a radius of 1.0 cm and a length of 50 cm. A current of 0.25 A flows through it. Use Ampere’s law to derive the formula for the magnetic field inside the solenoid (approximating it as an infinite solenoid). Then calculate the energy stored in the magnetic field.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:03 So here we are given a solenoid that is having thousand tons so i mentioned here its radius is one centimeter the length is 50 centimeters and the current flowing through it is 0 .25 ampers right now first you have to calculate the expression of magnetic field inside this solenoid using ampiers circuit law, right? so this law is actually the close integral or line integral of b .dor d l that is equal to mu not times the i that is enclosed right the current that is enclosed inside this loop so first you have to draw a loop right i've drawn it if you see here abc is a loop and we have to take the summation of this line integral about all these parts ab b c cd and da that is what i've did here okay so b times the magnetic field about ab right which is inside the solenoid times the length length okay so length i've taken as l on one side and l dash on the other side about this rectangle right similarly i've taken the line integral of all these sides right now the magnetic field inside the solenoid is somewhat like this right okay it will be through it moving through it in the axial direction right that means we'll have only magnetic field about abu itself in bc about this side and this side it will be zero there is no magnet field so we can consider this zero and da that is equal to zero sorry okay next about cd right cd cd is outside this solenoid itself so magnetic fill outside the soleno is also zero so these three terms has to be zero and if you see here that means we are left with integral of b a b times l that will be equal to mu not times the current that is enclosed in that loop right okay so b a b that is a magnet field that is produced that's what we need to calculate l is the length l mu not is you know and i enclosed if you see here the current that is enclosed is these are three currents that are entering or as per this let's say there are n coils so n currents are entering so that means we will have n times of i right right the number of turns number of current is about this loop right or in other words the magnitude field b will be equal mu not n times of l n by l times of i and this parameter is can be written as mu not n i n is the number of turns per unit length so this is the final expression of the magnetic field of this solenoid right then you have to calculate the energy that is stored in the magnetic field right so with that first we have to calculate this the value of magnetic field which is mu not which is 4x times 10 to the number of turns per unit length we have already have the number of turns is thousand and length is 50 centimeters right so here we have thousand by 50 centimeters which is 0 .5 meters times the current point to vibrate tesla so let's calculate and see it's 6 .28 times 10 to the bar of minus 4 tesla right so this the value of the magnetic field of this solenoid next you have to calculate the magnetic energy right so magnetic energy e will be equal to mu it's b square by mu not sorry it's b square by mu not times area of times the length right so b square we have 6 .28 times 10 to the power of minus 4 squared mu not is 4 pi times of 10 to the bar of minus 7 the area we are given area is the radius is given right which is equal to 0 .01 meters one centimeter okay so here will be pi r square so pi times of 0 .01 squared times the length is 50 centimeters which is 0 .5 meters and this will be in jules right so let's evaluate and see how much you're getting that's 4 .93 times 10 to the bar of minus 5 jules right so these are energy density sorry the energy that is stored in this solenoid...

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