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Yusf · Oct 23, 2016, 10:14 AM

The teacher at ilectureonline.com was deriving a formula for magnetic field produced by an infinite current carrying wire. Then he reached a stage like that shown at the attachment. Now, if I put infinity as the value of y, how does the part in brackets reduce to 1/R ?
He says r square plus infinity square equals infinity square. That I understand. So we have infinity square to the power of one half with equals infinity cubed.
Then one might say that infinity cubed equals infinity so the infinity on numerator cancels with that of the denominator. This sounds OK. But this will not work if y equals a very very large number.
So how does this integral work for a very very large value for upper limit of this integral? (PS the upper limit came from half the length of the wire.)

We know that the magnetic field near the centre of a very large wire should be very close to that created by an infinite wire.

ebaines · Oct 23, 2016, 12:22 PM

Please double check the equation - you wrote this for the part in the brackets:

$\frac {y}{R(R^2 + y^2)^{1.5}}$

(I assume that the exponents that look like letter v's are really 2's, based on your description.)

But I wonder if what you meant is this:

$\frac {y}{R(R^2 + y^2)^{\mathbf{1/2}}}$ ?

The integral as you wrote reduces to 0 for $y=\infty$ - it's essentially $\frac 1 {Ry^2}$ for y approaching $\infty$ . But if the exponent in the denominator is 1/2 instead of 1.5 the expression in the brackets approaches 1/R for large values of y.

Yusf · Oct 23, 2016, 03:00 PM

Yes I did make a mistake in my equation and I am extremely sorry for this. As I had a bad video quality, I mistook his "one half" as "one AND half". Reading your reply, I checked everything on online integral calculator and found the problem.
so yes that solves my problem. Thanks.
(never realised by squares look like v's . I will watch out for that from now on. )