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

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Vanya Saxena Grade: Upto college level

The current density acrossa cylindrical conductor of radius R varies according to the equation J=J0 [ 1-r/R], where r=distance from the axis. Thus the current density is maximum J0 at the axis r=0 and decreases linearly to zero at the surface r=R.Calculate the current in terms of J0 and the conductor's cross-sectional area A= pi R2 .


Supppose that instead the current density is a maximum J0 at the surface and decreases linearly to zero at the axis so that J=J0 r/R. Calculate the current.

8 years ago

Answers : (1)

AskiitianExpert Shine
10 Points


J=J0 [ 1-r/R]

      I= JdA from 0 to R

       dA=2pi*rdr considering it is a cylinder, so consider a ring element and hence the differntial area would come out to be this....

 On  integrating we get I=J0*pi*R2/3

Similarly integrate JdA for the other case to get I. ITs is exactly the same, just the J term needs to be changed in the integration.

8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details