Home
Class 12
PHYSICS
If vec(j) and vec(E ) are current and el...

If `vec(j)` and `vec(E )` are current and electric field inside a current carrying conductor at an instant then `vec(j). Vec( E )` is
(1) Positive (2) Negative
(3) Zero (4) May be positive or negative

Text Solution

AI Generated Solution

To solve the question regarding the scalar product of current density \(\vec{j}\) and electric field \(\vec{E}\) inside a current-carrying conductor, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding Current Density and Electric Field**: - Current density \(\vec{j}\) is defined as the current per unit area flowing through a conductor. It has a direction that is the same as that of the current. - The electric field \(\vec{E}\) inside the conductor is related to the potential difference and also points in the direction of the current flow. ...
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    AAKASH INSTITUTE ENGLISH|Exercise Try Yourself|34 Videos
  • CURRENT ELECTRICITY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT(SECTION-A(OBJECTIVE TYPE QUESTIONS))|69 Videos
  • COMMUNICATION SYSTEMS

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION D (Assertion-Reason)|10 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT (SECTION-D)|10 Videos

Similar Questions

Explore conceptually related problems

If vec(j) and vec(E ) are current density and electric field inside a current carrying conductor at an instant then vec(j). vec( E ) is (1) Positive (2) Negative (3) Zero (4) May be positive or negative

If vec(j) and vec(E ) are current density and electric field respectively inside a current carrying conductor , then correct relation is

Let vec(M) and vec(L) represent magnetic moment and angular momentum vectors for the electron in the above example. What is the sign of the dot product, vec(M).vec(L) . (positive, negative or zero) ?

If vec(E)_(1) and vec(E)_(2) are electric field at axial point and equatorial point of an electric dipole, then (1) vec(E)_(1).vec(E)_(2)gt 0 (2) vec(E)_(1).vec(E)_(2)=0 (3) vec(E)_(1).vec(E)_(2)lt 0 (4) vec(E)_(1)+vec(E)_(2)=vec(0)

Electric field on the axis of a small electric dipole at a distance r is vec(E)_(1) and vec(E)_(2) at a distance of 2r on a line of perpendicular bisector is

If vec a and vec b are non-zero and non-collinear vectors, then vec ax vec b=[ vec a vec b hat i] hat i+[ vec a vec b hat j] hat j+[ vec a vec b hat k] hat k vec adot vec b=( vec adot vec i)( vec adot hat i)( vec bdot hat j)+( vec adot hat j) ( vec bdot hat j) + ( vec adot hat k)( vec bdot hat k) If vec u= hat a-( hat adot hat b) hat b and hat v= hat ax hat b , then | vec v|=| vec u| If vec c= vec ax( vec ax vec b) , then vec c dot vec a=0

The displacement of a charge Q in the electric field vec(E) = e_(1)hat(i) + e_(2)hat(j) + e_(3) hat(k) is vec(r) = a hat(i) + b hat(j) . The work done is

If vec a= i- j and vec b=- j+2k ,find (vec a-2 vec b)dot( vec a+ vec b)

The potential field of an electric field vec(E)=(y hat(i)+x hat(j)) is

The potential field of an electric field vec(E)=(y hat(i)+x hat(j)) is