#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# A loThe current density across a cylinder conductor of radius R

Susmita
425 Points
2 years ago
You have to use polar coordonate (r,$\theta$).
An elemental area dS=rdrd$\theta$.
Now elementary current is
dI=J.dS
We have to integrate r from 0 to R,$\theta$ from 0 to 2$\pi$ to obtain total current I.
$\int_{0}^{R} Jo (1-\frac{r}{R})rdr\int_{0}^{2\pi}d\theta$
Please carry it out yourself. If you face problem then only ask me.
The ans will be $(\pi Jo R^2)/3$
The next problem will be in the same manner with the value of J as given in it.