Mechanics or what? - 5

Electricity and Magnetism Level 4

A conducting rod of mass $m$ and length $L$ is free to slide on smooth vertical conducting rails as shown. Capacitance of the capacitor is C. Uniform magnetic field intensity $B$ exists through out the region and is normally inward as shown. The rod is released from rest at time $t=0$ . Find the velocity of the rod as a function of time.

Assume that gravitational force is more than magnetic force

Try my previous problem here

v=\dfrac{mgt}{B^2L^2C}e^{\frac{mgC}{B^2L^2}}

None

v=\dfrac{mgt}{B^2L^2C+m}

\dfrac{B^2L^2Cgt+2m}{2m}

\dfrac{B^2L^2Cgt}{2m}

v=gt

1 solution

Alan Enrique Ontiveros Salazar
Jul 2, 2016

Let $v$ be the velocity at any point of the movement of the rod. We see that the induced emf is $\epsilon=BLv$ . In this circuit, the current in terms of time is $I=C\dfrac{\mathrm d\epsilon}{\mathrm dt}=BLC\dfrac{\mathrm dv}{\mathrm dt}$ .

The magnetic force, which is directed upwards, has a magnitude of $F_m=ILB=B^2L^2C\dfrac{\mathrm dv}{\mathrm dt}$ .

Applying Newton's second law we get:

$ma=mg-F_m \implies m\dfrac{\mathrm dv}{\mathrm dt}=mg-B^2L^2C\dfrac{\mathrm dv}{\mathrm dt}$

$(m+B^2L^2C)\mathrm dv=mg\,\mathrm dt$

Integrate both sides using the initial conditions:

$\displaystyle (m+B^2L^2C)\int_{0}^{v} \mathrm dv=mg\int_{0}^{t}\mathrm dt$

$(m+B^2L^2C)v=mgt \implies \boxed{v=\dfrac{mgt}{m+B^2L^2C}}$

Mechanics or what? - 5

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