Playing in an Ampere-ian World!

Electricity and Magnetism Level 5

There is a circular current carrying loop of radius R R , it has a current i i passing through it, and its North (anti-clockwise face) is facing upward

Suppose, i place a thin disc magnet of mass m m with negligible dimensions over it that has a Magnetic moment of M M , such that the north pole is facing downwards at height d d above the loop.

Derive an expression for the force of interaction In Newtons between them, and submit the value for the given details.

Details and assumptions

  • m = m= 1 g
  • M = 20 M=20 Am 2 ^2
  • R = R = 10 cm
  • d = 20 d= 20 cm
  • i = 1 i=1 A
  • Take the magnetic permeability for vacuum to be 4 π × 1 0 − 7 4\pi\times10^{-7}

Note, to derive the expression, you may use the model of magnetic charges, also neglect negligible terms, I assure you that all information you need has been given.


The answer is 0.0001348.

Mvs Saketh by Mvs Saketh , 24, India

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1 solution

You can solve this problem in 3 way i will introduce each of them.

  1. The most easy way:

For axial-symmetry field you can easy derive that the force acting on a dipole moment p m p_m is F = − p m d B d z F=-p_m \frac{dB}{dz} Putting B = μ 0 i 2 R 2 ( R 2 + z 2 ) 3 2 B=\frac{\mu_0 i}{2}\frac{R^2}{(R^2+z^2)^{\frac{3}{2}}}

We will find that F ( z = d ) = 3 2 p m μ 0 i R 2 d ( R 2 + d 2 ) 5 2 F(z=d)=\frac{3}{2} p_m \mu_0 i \frac{R^2 d}{(R^2+d^2)^{\frac{5}{2}}}

  1. 'Creator's' way

If you interpret dipole as 2 monopole of magnetic charge q m q_m and distance a a then p m = q m a p_m=q_m a . Using the fact that force is equal to F ( z ) = q m B ( z ) F(z)=q_m B(z) and again you will get

F ( z = d ) = 3 2 p m μ 0 i R 2 d ( R 2 + d 2 ) 5 2 F(z=d)=\frac{3}{2} p_m \mu_0 i \frac{R^2 d}{(R^2+d^2)^{\frac{5}{2}}}

3.Third Newton's law

You can first find the force of magnetic moment using the formula for field of magnetic moment and then by third Newton's law the force is same and again you get

F ( z = d ) = 3 2 p m μ 0 i R 2 d ( R 2 + d 2 ) 5 2 F(z=d)=\frac{3}{2} p_m \mu_0 i \frac{R^2 d}{(R^2+d^2)^{\frac{5}{2}}}

Note: You can apply third Newton's law on magnetic force only if two magnetic moments as a vector are parallel!

So the correct answer is

F ( z = d ) = 3 2 p m μ 0 i R 2 d ( R 2 + d 2 ) 5 2 = 1.348 ˙ 1 0 − 4 N F(z=d)=\frac{3}{2} p_m \mu_0 i \frac{R^2 d}{(R^2+d^2)^{\frac{5}{2}}}=1.348\dot{}10^{-4} N

3 Helpful 0 Interesting 0 Brilliant 0 Confused

This is not a solution. It's just cutting and pasting the formula and plugging in the values. You haven't derived anything. Write the complete complete solutionimstead of just saying that it is easy to derive.

raj abhinav - 1 year, 3 months ago

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