What is the fraction of work per second by F is converted into heat.

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Two long parallel horizontal rails, a distance l apart and each
has a resistance $\lambda$ per unit length are joined at one end by a
resistance R. A perfectly conducting rod MN of mass m is free to
slide along the rails without friction. There is a uniform magnetic
field of induction B normal to the plane of paper and directed
into the paper. A variable force F is applied to the rod MN such
that, as the rod moves, a constant current i flows through R.
What is the fraction of work per second by F is converted into heat.

edited Feb 20
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1 vote

Your equation #1 is $$(R+2\lambda x)i=B\ell v=B\ell \frac{dx}{dt}$$ Solving for $x$ as a function of time, and assuming $x(0)=0$, we get $$R+2\lambda x(t)=R e^{kt}$$ where $k=\frac{2\lambda i}{B\ell}$. The velocity of the bar is $v(t)=v_0e^{kt}$ where the initial velocity is $v_0=\frac{iR}{B\ell}$.

We do not need to find an expression for $F$. And we do not need to take account of a resistive force $Bi\ell$ which the magnetic field exerts on the wire. This is because the motion of the wire in the magnetic field is the cause of the current. The magnetic field does not create a current then exert a force on the same current which it has created. If the current had been generated independently of the magnetic field, then there would be a magnetic force on that current.

No work is done by the magnetic field. Only force $F$ does work. The rate of work $P$ done by force $F$ is the sum of 3 components :
(1) the rate $K$ at which the kinetic energy of the bar is increasing;
(2) the power $Q$ dissipated as heat in the resistors; and
(3) the rate $M$ at which energy is being stored in the magnetic field created by the rectangular loop of current.

Now $M$ depends on the self-inductance $L(x)$ of the rectangular loop, which depends on $x$. Even for such a simple geometry an expression for $L(x)$ is difficult to obtain - see Rectangle Loop Inductance Calculator in All About Circuits website. I shall assume without justification that $M \ll P$ is negligible.

$$K=\frac{d}{dt}(\frac12mv^2)=mv\frac{dv}{dt}=mkv^2=mkv_0^2 e^{2kt}$$ $$Q=i^2(R+2\lambda x)=i^2Re^{kt}$$

The fraction of work done by $F$ which is converted into heat is $$\frac{Q}{P}=\frac{Q}{Q+K}=\frac{1}{1+K/Q}=\frac{1}{1+he^{kt}}$$ where $$h=\frac{mkv_0^2}{i^2 R}=\frac{2\lambda m i R}{(B\ell)^3}$$ This is not constant, it decreases with time. The initial fraction at $t=0$ is $\frac{1}{1+h}$. If the rails have zero resistance ($\lambda=0$) then $k=K=h=0$ so $Q=P$ - ie all of the work done by $F$ is dissipated as heat in resistor $R$, as obtained in Q4 in this worksheet.

answered Feb 22 by (27,948 points)
edited Feb 23