Biot Savart law (vector form) | Moving charges & magnetism | Khan Academy

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[@lukeverst3653]

This is exactly what I needed! Thank you! Explained it very well.

[@user-cc7ks7re3w]

Sir one doubt, in the textbook and other sources the vector form of Biot-Savart law has r^(3) but yours is r^(2)... Can you please clear my doubt??

[@felixwegmuller351]

You have to use r^3 if you dont use the unit vector. the guy used unit vectors which is why he uses r^2. for more info you can look at the english wikipedia article about biot-savart

[@user-cc7ks7re3w]

@@felixwegmuller351 Thank you ☺️

[@oreowithurea5018]

@@user-cc7ks7re3w the other explained it pretty well, you can either use unit vector in the direction of radius to denote the vector for radius, or just simplify the unit vector which turns out to be *r* /r , plug this value instead of unit vector and you'll get the textbook equation

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For refreshment🤞

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Literally

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blud studying this in college? this is highschool physics

[@rizzwan-42069]

@@Devilhunter69 college tends to deepen what you learn in highschool in my experience. well actually i'm in university but anyway. i'm doing computer science so the physics level is closer to high school level but still bit higher than what i learned in highschool. it uses basically the same material but has more explanation and where the formulas are from and uses more calculus.

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[@mermelada.]

Hermosa explicación, gracias !

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your handwriting

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For each tiny segment, do we have to do a vector sum?

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Very clear explanation. Thank you!

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[@krasimirronkov17]

Can you make a video explaining capacitors

[@kaushikb2819]

Hey here's a doubt, our NCERT books say idl*r^ ^3 but u hv written r^2 instead of r^3 so which one's right?

[@rinzler3202]

hello kaushik rcap can also be said as r vector . r = so it becomes rvector / r^3

[@hemavathy9806]

Did you mean by substituting r cap=(r vector/magnitude of r vector)?

[@rinzler3202]

@@hemavathy9806 es

[@hemavathy9806]

👍

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@@rinzler3202 ahh I see...thank you 😀

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[@allison7612]

Isn’t this finding just the magnetic field as it is affected by the length of dl? So to find the total force applied on a single point you would need to do multiple equations? Or should you just use the perpendicular measurement (since it is where force is strongest) for an estimate?

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It helped a lot. Thank u

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[@cringenatao7158]

Thank you, Mahesh, very cool

[@VincentAlesi]

Really nice explanation

[@chetanganachari4882]

Sir in NCERT it is given that in vector form the distance is r^3 but you mentioned it as r^2...please clarify it

[@TripuriansYT]

We know r is a unit vector So, r^ = r(vector)/ |r| Or, r^ = r(vector)/r Since, |r| = r So, put the value of r^ then you get r³

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@@tasneem6427 Around 4-8 hours! :)

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This was awesome!

[@iqraafzal517]

I appreciate your efforts. But you lacked in explaining why magnetic field changes when we change angle. You just explained the effect but skipped the cause...!

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Watch previous videos.

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[@crossmindzgaming7998]

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[@namenotfound1118]

why only dl vector cross r vector why not r vector cross dl vector. Could anyone explain or which one of your videos explain this?

[@whitedaisy2263]

Concept: Vectors - Cross Product Suppose we have two vectors, A=2i+3j and B=4i−j where i and j are unit vectors along the x and y axes, respectively. The cross product A×B is given by: A×B= ∣ i j k ∣ 2 3 0 4 -1 0 Now, calculating the cross product; A×B=(3×0−(−1)×0)i−(2×0−4×0)j+(2×(−1)−3×4)k we get: AxB =0i-0j-12k A×B= −12k. [It is in the negative z- axis] But in this case if we do B x A: By doing the same method again as shown above but putting values accordingly. Mathematically the relationship is expressed as: B x A = - ( A x B ) A x B= -12 (as calculated) B X A = - (-12k) = +12k [it is in Positive Z axis] Difference in the direction!! Similarly, we cannot write r x dl.

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[@catherinefernandez9654]

Why is it r^3 in my textbook not r^2..

[@joenuts3308]

In his equation, he has taken the unit vector of r that is r cap. But in textbook it is vector r which is equal to the unit vector times the magnitude of r so we have to multiply r on numerator and denominator to get db = μ0/4π x I dl × vector r/r^3

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God bless you Sir ❤

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[@ShashiSharma-uv2vr]

Why you take DL as a vector??? (Length is scalar) 😅

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Beo savar law😳

[@demonking2508]

shouldnt it be r3 up

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[@aniketprasad7700]

Hey in ncert textbooks we have inversely proportional to r^3 relation but here it is r^2. also your calculation from solving vector equating down to magnitude equation is wrong

[@abhinavsrivastava7157]

Notice that in ncert the cross product is between dl and r VECTOR(and not r cap)... r vector is essentially the magnitude of r multiplied by unit vector of r(r cap) Therefore, dB = I dl × r / r³ = I dl × mag(r).r cap/r³ The magnitudes in the numerator and denominator gets cancelled and we're left with dB = I dl r cap / r² both derivations are equivalent. You can use either Hope this helps

[@sumitkumarsingh5290]

Why (dl × r ).not (r × dl)

[@milandas2128]

Sorry, i do not know in detail 😅 . What i know is, in cross product, the order matters. The direction of the magnetic field will change if u take (r x dl) instead of (dl x r). The direction will be in accordance with the "right hand rule". if u take (r x dl), your magnitude will be the same, but your direction will be opposite to that of the right hand rule. Thats what i know😅.

[@srinivaskotti1981]

@@milandas2128 thank you

[@rixoncusher]

0:01

[@saurabhkapse697]

In India we pronounced as it is given in book like biot savart law 😅😅