Question
When a current in a circular loop is equivalently replaced by a magnetic dipole:
  1. The pole strength m of each pole is fixed.
  2. The distance d between the poles is fixed.
  3. The product md is fixed.
  4. None of the above.

Answer

  1. The product md is fixed.

Explanation:

When we replace a circular current-carrying loop with a magnetic dipole to resemble field lines of the circular loop, the pole strength m and the distance between the poles are not fixed.

But the magnetic dipole moment of both systems is always fixed. It is the product of the magnetic moment and the distance between the poles. In other words, md is fixed.

A current loop of area A and current I can be replaced with a magnetic dipole of dipole moment md.

i.e. md = IA

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

A potentiometer consists of a wire of length 4 m and resistance 10 W. It is connected to cell of emf 2 V. The potential difference per unit length of the wire will be

(a) 0.5 V/m

(b) 10 V/m

(c) 2 V/m

(d) 5 V/m

A varying current at the rate of 3 A/s in a coil generates an e.m.f. of 8 mV in a nearby coil. The mutual inductance of the two coils is

(a) 2.66 mH

(b) 2.66  

(c) 2.66 H

(d) 0.266 H

The mutual inductance between a primary and secondary circuits is 0.5 H. The resistances of the primary and the secondary circuits are 20 ohms and 5 ohms respectively. To generate a current of 0.4 A in the secondary, current in the primary must be changed at the rate of

(a) 4.0 A/s

(b) 16.0 A/s

(c) 1.6 A/s

(d) 8.0 A/s

A conducting disc of radius r spins about its axis with an angular velocity ω. There is a uniform magnetic field of magnetude B perpendicular to the plane of the disc. C is the centre of the ring.

In the circuit shown here, the readings of the ammeter and voltmeter are

(a) 6 A, 60 V

(b) 0.6 A, 6 V

(c) 6/11 A, 60/11 V

(d) 11/6 A, 11/60 V

The electric field E, current density J and conductivity s  of a conductor are related as

(a) σ = E/j  

(b) σ = j/E

(c) σ = jE

(d) σ = 1/jE

There will be no force between two wires carrying currents if currents are:
If work function is $\phi$, then the formula for threshold wavelength is :
When the length of the tube of a microscope is increased, the magnifying power will be :
The critical mass of a fissionable material is: