Home
Class 11
PHYSICS
A uniformly charged disc of radius r and...

A uniformly charged disc of radius r and having charge q rotates with constant angular velocity `omega`. The magnetic dipole moment of this disc is `(1)/(n)qomegar^(2)`. Find the value of n.

Text Solution

AI Generated Solution

The correct Answer is:
To find the value of \( n \) in the expression for the magnetic dipole moment of a uniformly charged rotating disc, we can follow these steps: ### Step 1: Understand the problem We have a uniformly charged disc of radius \( r \) with total charge \( q \) rotating with a constant angular velocity \( \omega \). We need to find the magnetic dipole moment \( \mu \) and relate it to the given expression \( \frac{1}{n} q \omega r^2 \). ### Step 2: Recall the formula for magnetic dipole moment The magnetic dipole moment \( \mu \) for a rotating charge distribution can be expressed in terms of angular momentum \( L \) and total charge \( q \): \[ \mu = \frac{L \cdot q}{2m} \] where \( m \) is the mass of the object. ### Step 3: Calculate the angular momentum \( L \) The angular momentum \( L \) of the disc can be calculated using the moment of inertia \( I \) and the angular velocity \( \omega \): \[ L = I \omega \] For a disc, the moment of inertia \( I \) is given by: \[ I = \frac{1}{2} m r^2 \] Thus, substituting this into the expression for angular momentum: \[ L = \left(\frac{1}{2} m r^2\right) \omega \] ### Step 4: Substitute \( L \) into the magnetic dipole moment formula Now substituting \( L \) into the formula for \( \mu \): \[ \mu = \frac{\left(\frac{1}{2} m r^2 \omega\right) q}{2m} \] Simplifying this gives: \[ \mu = \frac{q r^2 \omega}{4} \] ### Step 5: Compare with the given expression We are given that: \[ \mu = \frac{1}{n} q \omega r^2 \] Now we can set the two expressions for \( \mu \) equal to each other: \[ \frac{q r^2 \omega}{4} = \frac{1}{n} q \omega r^2 \] ### Step 6: Solve for \( n \) Since \( q \), \( \omega \), and \( r^2 \) are common on both sides, we can cancel them out (assuming they are not zero): \[ \frac{1}{4} = \frac{1}{n} \] Taking the reciprocal gives: \[ n = 4 \] ### Final Answer Thus, the value of \( n \) is \( 4 \). ---
Promotional Banner

Topper's Solved these Questions

  • MAGNETIC EFFECT OF CURRENT AND MAGNETISM

    DC PANDEY ENGLISH|Exercise Matrix Matching type Questions|12 Videos
  • LAWS OF THERMODYNAMICS

    DC PANDEY ENGLISH|Exercise Level 2 Subjective|18 Videos
  • MEASUREMENT AND ERRORS

    DC PANDEY ENGLISH|Exercise Subjective|19 Videos

Similar Questions

Explore conceptually related problems

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.

A uniformly charged solid sphere a mass M, radius R having charge Q is rotated about its diameter with frequency f_0 . The magnetic moment of the sphere is

A nonconducing disc of radius R is rotaing about axis passing through its cente and perpendicual its plance with an agular velocity omega . The magnt moment of this disc is .

A thin uniform ring of radius R carrying charge q and mass m rotates about its axis with angular velocity omega . Find the ratio of its magnetic moment and angular momentum.

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 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.

Solid uniform conductiong sphere of mass 'm' and charge Q , rotates about its axis of symmetry with constant angular velocity 'omega' then the ratio of magnetic moment to the moment of inertia of the sphere is (xQomega)/(6m) then x is :(Neglect induced charges due to centrifugal force)

A disc of radius R rotates with constant angular velocity omega about its own axis. Surface charge density of this disc varies as sigma = alphar^(2) , where r is the distance from the centre of disc. Determine the magnetic field intensity at the centre of disc.

A ring of mass m and radius R is given a charge q. It is then rotated about its axis with angular velocity omega .Find (iii)Magnetic moment of ring.