IIT JEE , AIEEE Forum - Physics , Magnetism

ramyani

Q. A hypothetical magnetic field existing in a region is given by B = B ₀ e _r where e _r denotes the unit vector along the radial direction. A circular loop of radius a , carrying current , is placed with its plane || to the X - Y plane and the centre is at ( 0 , 0 , d ). Find the magnitude of the magnetic force acting on the loop.

Salute assured for detailed, repeat, detailed solution.

anchitsaini

I hope the figure is self explanatory :)

$\mbox{Considering an element of length dl as shown } \\ \\<br/>dF = i dl B_{0} \sin \theta \\ \\<br/>F = i * 2 \pi a B_0 \sin \theta \\ \\<br/>= \frac {2 \pi i B_0 a^2}{\sqrt{d^2 + a^2}}$

ramyani

no i don't understand.

1. magnetic field along the radial direction => along the radius of the loop.

2.In that case the angle b/w B ( along the radius ) and dl is 90 degree.

Conjurer

Nah, radial always doesnt mean along the radius of a circle. Don't you say force on a charge q on another charge q is radial?

So I guess now you understand, I also had a doubt in this.

ramyani

I still don't understand.

ramyani

why this dia ? why not other type of diagram ?

Do we hav to do all of hcv magnetics ?

feynmann

It is simple !!!!!!!!!!!!

I am giving u the formal approach to solve the problem

see , dF = I dl * B

Now B = B er

& dl = dl e ( Where er ,e& e are the unit vectors in the spherical coordinate )

Taking the cross product , we have dF = I B e dl

( because e * er = e )

Now e = cos

cos

i + cos

sin

j - sin

k

& dl = a d

Integrating dF , (

limiting from 0 to 2

) noting that i & j component cancels ,

We have F = 2

BIa sin

k

Now sin

= a / sqrt ( a^2 + d^2 )

so Force = F = 2

BI a^2 / sqrt ( a^2 + d^2 ) k

feynmann

But I should warn you that this is not a physical problem , only a mathematical exercise .

Because , according to Maxwell's eqn we must have

div B = 0 everywhere in space for all time !!

But for B = B er we have div B = 2 B / r which is non zero everywhere .

So nowhere in our universe you can find such a magnetic - field distribution

ramyani

oh richard !

u r a genius.

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