dear students , now we are going to discuss

another important law, in magnetic field calculation that is ampere’s law, like , we have studied

in electrostatics . we have studied that in electrostatics. Goss law. is easier , in calculation

. of electric field for symmetric charge distributions. because we have studied that in non uniform

charge distribution gauss law cannot be directly , applied for easy calculations . but in case

of symmetric charge distribution using gauss law we can easily calculate the electric field

strength . similar to this, we can write . for . magnetic

induction calculations , for , symmetric distribution of currents- or symmetric current distribution

, amp ere’s law . is helpful , in making. calculation part, easy, calculations easier,

so we will discuss am pair’s law somehow we can say it is similar to gauss law but

it is slightly different because . goss law was giving us , the surface integral for closed

surface whereas in amp ere’s law we calculate the line integral , of magnetic induction

for a closed path. so we will discuss about amp ere’s law , let’s

continue it. let’s state the am pair’s law , for which

we can write , amp ere’s law . is stated as , it is written as the line integral of,

magnetic induction, along. any closed path. in a region, is equal to . the mu knot times-

which is the magnetic permeability of free space it is mu knot times , the total current

, crossing, the area enclosed, by that closed path. this statement we call as amp ere’s

law. in this situation , say in a region there

are current carrying wires carrying a current i1 , this is carrying current i2 , this is

another wire which is carrying opposite current i3 , in this situation say we consider a closed

path like this, which is enclosing the 3 currents . and say the direction of path we have taken

as anti clockwise and the name of path is m.

then , to find the line integral of magnetic induction we need to consider a small element

in this path say the element is taken as dl . and at the location of dl due to these wires

, as well as due to another wires which are located outside the region , say this is the

wire i4, this is the wire i5 and so on, the net magnetic induction at the location of

dl is given by b-vector ,then here we can write integration of b dot dl that is line

integral of magnetic induction. at , a point on, this path m , and say if

this is integrated for a closed path m, then according to amp ere’s law this line integral

for the whole closed path is written as mu knot times the whole current enclosed by,

the area, through which the current is passing , in this situation we can see, the total

current passing through this area will be i1- i2 and minus i3 .

so this can be written as i1+ i2 minus i3 , in this situation there is an important

point if we consider that in which direction we should take current , so here we can write

that, direction of current, is taken positive , according to right hand thumb rule . applied,

on the closed path , in this situation if this closed path is taken anti clockwise if

we curl our right hand fingers , along the closed path then the direction of your thumb.

will be the direction of positive current so here, we have taken i1 and i2 positive

current and i3 is taken as negative current . and this amp ere law can be re-written as

integration for the closed path, b dot dl , we can write as , mu knot multiplied by

current enclosed , this is, the expression for the am pair’s law , which we will use

, in upcoming applications, we must be careful about this.

how do we place the amperian loop so that magnetic field vanishes

plz answer

very nice superbbbbbb sir

Thank you very much

Respected Sir, if we consider a finite current carrying wire and if we want to calculate magnetic field at a perpendicular distance 'r' from its 'midpoint' using amperes law, then if i consider a circular loop and calculate, I am not getting the right ans. Please tell me Sir where I am wrong..

Sir,

When we are using Amperes Law….what do we do if there is a current along the plain of the loop… do we consider it or do we say that it's zero??

sir, you have considered the wires outside the loop also but in many books they have strictly not considered the wires outside the loops which is correct?

sir is this law like Gauss law where only enclosed current is taken into consideration but effect (integration of b.da ) is due to all present current in the surrounding??

sir then whaat abt currents outside have no effect

thankyou sir you explain so well

sir i have studied somewhere that we can not use this law in case of variable electric flux and to apply law in this case we add displacement current sir what is the meaning of this ?..

Sir,does this outside wires contribute to net magnetic field inside the integral……or..we have to consider only inside wires magnetic fields. ….?

Sir why is this not applicable for a finite wire

Great explanation. Hope you made bigger videos rather than smaller multiple vids sir.