Question

e Iz OI, I5 5 I. de AZ (5) What does Ampere's law state regarding the closed loop path integral ¢ B. dl around the loop shown in terms of the currents Ii to Is shown in the first figure →? Discuss briefly. A thick straight conducting pipe (see second figure of radius R has a non-uniform current 1. (A) flowing through it in the axial direction 2 . The current density is distributed as a function of r as J(r) = kr’ î where k is a constant. (i) Calculate the total current lo in terms of k and R. (ii) What now is k in terms of lo and R? (iii) Calculate the B-field as a function of r for r s R in terms of lo not k. (iv) Calculate the B-field as a function of r for r R in terms of Io. (v) Make up a clear graphical sketch of the magnitude of B vs. r and label the maximum value of B at r = R and how B rises and falls as a function of r for all r. is 14 IR

Answer 1

According to Ampere's Circuital Law : The line integral of net magnetic field along a closed loop is equal to net current enclosed in the loop multiplied by $\mu_{o}$ .

Here Net magnetic field is not only due to current within the loop, it is due to all the currents inside and outside the loop.

So, in the diagram shown the magnetic field is due to all the currents present, .

and the net current enclosed depends on the direction of current like here, net current enclosed is .

Now, see the diagram :

E A SR 3

i) consider a small circular loop of radius r and thickness dr .

the current in the element is $di = J(2\pi r dr)$

the total current is $I_{o} = \int_{0}^{R}J(2\pi r dr) = \int_{0}^{R}kr^{3}(2\pi r dr) = 2\pi k\int_{0}^{R}r^{4}dr = \frac{2\pi kR^{5}}{5} A$ [answer]

ii) the value of k = $\frac{5I_{o}}{2\pi R^{5}}$ [answer]

iii) Consider a loop A, of radius r <= R :

current enclosed in the loop is $i_{enc} = \frac{2\pi kr^{5}}{5} A$

applying Ampere's Law :

$B*2\pi r = \mu_{o}*\frac{2\pi kr^{5}}{5}$

$\Rightarrow B = \frac{\mu_{o} kr^{4}}{5} T$

iv) cosinder a loop B of radius r > = R.

current enclosed in the loop is $i_{enc} = I_{o} = \frac{2\pi kR^{5}}{5} A$

applying Ampere's Law :

$B*2\pi r = \mu_{o}*I_{o}$

$\Rightarrow B = \frac{\mu_{o}I_{o}}{2\pi r} T$

v) see the graph :

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