Home
Class 12
PHYSICS
A non-conducting ring of radius R has ch...

A non-conducting ring of radius R has charge Q distributed unevenly over it. If it rotates with an angular velocity `omega` the equivalent current will be:

A

zero

B

`Qomega`

C

`Qomega/(2pi)`

D

`Qomega/(2pir)`

Text Solution

Verified by Experts

The correct Answer is:
c
Rotating charge is equivalent to current
`I=Q/T` where=T =time period `=(2ip)/omega`, Hence `I=(Q omega)/(2 pi)`
Promotional Banner

Topper's Solved these Questions

  • DC CIRCUIT

    VMC MODULES ENGLISH|Exercise LEVEL-2 DAILY TUTORIAL SHEET-2|10 Videos

  • DC CIRCUIT

    VMC MODULES ENGLISH|Exercise LEVEL-2 DAILY TUTORIAL SHEET-3|10 Videos

  • DC CIRCUIT

    VMC MODULES ENGLISH|Exercise LEVEL-1 DAILY TUTORIAL SHEET -5|15 Videos

  • CURRENT ELECTRICITY

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE-F|10 Videos

  • DYNAMICS OF A PARTICLE

    VMC MODULES ENGLISH|Exercise JEE Advance (Archive) Level - II (SINGLE OPTION CORRECT TYPE )|31 Videos

Similar Questions

Explore conceptually related problems

A conducting sphere of radius r has a charge . Then .

As shown in figure there is a ring having radius R and charge q distributed uniformly over it. If ring is rotated with a constant angular velocity @, then torque acting on the ring due to magnetic force is (Assume that magnetic field B is parallel to plane of ring)

A thin disc of radius R and mass M has charge q uniformly distributed on it. It rotates with angular velocity omega . The ratio of magnetic moment and angular momentum for the disc is

A non-conducting thin disc of radius R charged uniformly over one side with surface density s rotates about its axis with an angular velocity omega . Find (a) the magnetic induction at the centre of the disc, (b) the magnetic moment of the disc.

A closed current-carrying loop having a current I is having area A. Magnetic moment of this loop is defined as vec(mu) = vec(IA) where direction of area vector is towards the observer if current is flowing in anticlockwise direction with respect to the observer. If this loop is placed in a uniform magnetic field vec(B) , then torque acting on the loop is given by vec(tau) = vec(mu) xx vec(B) . Now answer the following questions: Let ring in the above question is having a radius R and a charge Q is uniformly distributed over it. Ring is rotated with a constant angular velocity (omega) as mentioned above. Torque acting on the ring due to magnetic force is

A disc of mass m has a charge Q distributed on its surface. It is rotating about an XX' with a angular velocity omega . The force acting on the disc is

A conducting ring of radius r having charge q is rotating with angular velocity omega about its axes. Find the magnetic field at the centre of the ring.

A conducting ring of radius r having charge q is rotating with angular velocity omega about its axes. Find the magnetic field at the centre of the ring.

consider a nonconducting ring of radius r and mass m which has a total charge q distributed uniformly on it the ring is rotated about its axis with an angular speed omega . (a) Find the equivalent electric current in the ring (b) find the magnetic moment mu of the . (c) show that mu=(q)/(2m) where l is the angular momentum of the ring about its axis of rotation.

A non-conducting disc having unifrom positive charge Q , is rotating about its axis with unifrom angular velocity omega .The magnetic field at the centre of the disc is.