Mechanics or what? - 2

Electricity and Magnetism Level 3

A horizontal metallic rod of resistance $R$ and of mass $m$ is free to slide on two vertical conducting rails as shown. The terminal speed of the rod is

Details and Assumptions:

Uniform magnetic field intensity $B$ exist throughout the region and is normally inward.
The length of the rod is $L$ .
The inductance of the inductor is $M$ .

Try my previous problem here

\dfrac{mgR}{B^2L^2}+M

\dfrac{mgM}{RB^2L^2}

\dfrac{mgR}{B^2L^2}

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

\dfrac{5mgR}{7B^2L^2}

1 solution

Hjalmar Orellana Soto
Jun 27, 2016

By Kirchhoff's we have that $IR+M\frac{dI}{dt}=BLv$ at terminal velocity. The differential equation can be solved like this $M\dot{I}+RI=BLv$ $e^{\frac{R}{M}t}\dot{I}+e^{\frac{R}{M}t}\frac{R}{M}I=e^{\frac{R}{M}t}\frac{BLv}{M}$ $\frac{d}{dt}(e^{\frac{R}{M}t}I)=e^{\frac{R}{M}t}\frac{BLv}{M}$ $e^{\frac{R}{M}t}I=e^{\frac{R}{M}t}\frac{BLv}{R}$ $I=\frac{BLv}{R}$

As at this moment both of the forces are equal, we have: $mg=I\displaystyle \oint d\vec{l} \times \vec{B}$ $mg=\frac{B^2L^2v}{R}$ $v=\frac{mgR}{B^2L^2}$

We can directly do it like this - when terminal velocity is reached , force = mg (constant) that implies (I) is constant , hence di/dt is zero . that saves time.

In this q , we don't require solving differential equation.

Aniket Sanghi - 4 years, 8 months ago

Mechanics or what? - 2

Try my previous problem here

1 solution

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