0
$\begingroup$

I came across a question regarding the force between two magnetic dipoles $M_1$ and $M_2$ separated by a distance $x$ .

Here in this text book I am given the solution is

$$ B = \cfrac{\mu_0}{4 \pi} \cfrac{2M_1}{x^3}$$ which on differentiating with respect to distance $x$ gives

$$\frac{dB}{dx} =\cfrac{\mu_0}{4 \pi} \cfrac{6M_1}{x^4} $$ Further,

$$ F = -M_2 \frac{dB}{dx} = \cfrac{\mu_0}{4 \pi} \cfrac{6M_1 M_2}{x^4} $$

What is correct explanation of relation of '$\frac{dB}{dx}$' with force here?

$\endgroup$
2
$\begingroup$

In school we learn that energy (or work) is given by $E = -\int F \cdot dx$. Hence, we can write $dE = -F \cdot dx$. From here you see, that $F = -\frac{dE}{dx}$. All you have to do is to compare your equation with mine, $dE = M \cdot dB$, and you realise, that the magnetic dipole moment times the B-field is an energy. The term $dB/dx$ is the "gradient of the magnetic field" (first derivative).

$\endgroup$
  • $\begingroup$ I think you should say $dE=MdB$. You can't have a differential in only one side of the equation. $\endgroup$ – Ballanzor Jun 15 at 11:09
  • $\begingroup$ Thank you. I edit it. $\endgroup$ – Semoi Jun 15 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.