Terminal velocity

Electricity and Magnetism Level 3

A system of long, parallel conducting rails is set up in the vertical plane and joined at both the ends by resistors as shown in the figure. A uniform magnetic field exists perpendicular to the plane of the rails denoted by $\vec{B}$ . A rod whose length is equal to the separation between the rails is allowed to fall freely and attain a terminal velocity. If the powers dissipated by the resistors $R_1$ and $R_2$ at steady state are $0.76 \text{ W}$ and $1.2\text{ W},$ respectively, then what is the terminal velocity of the rod?

Details and Assumptions

The length and mass of the rod are 1 m and 0.2 kg.
The acceleration due to gravity is $9.8\text{ m/s}^2$ .
The magnitude of the magnetic field is 0.6 T.
The rails are long enough so that the steady state conditions can be easily analysed.

The answer is 1.

3 solutions

Chew-Seong Cheong
Aug 18, 2014

At steady state, by conservation of energy the rate of work done by the falling rod $\frac{dW}{dt}=mgv_{terminal}$ is equal the power dissipated by the two resistors $R_1$ and $R_2$ , $P = 0.76+1.2 = 1.96W$ . Therefore:

$\cfrac{dW}{dt} = P$ $mgv_{terminal} = P$ $\Rightarrow v_{terminal} = \cfrac{P}{mg} = \cfrac{1.96}{0.2\times 9.8} = \boxed {1m/s}$

Jatin Yadav
Aug 17, 2014

Total power dissipated = $\epsilon I = BvLI = 1.96$

$\Rightarrow IBL = \dfrac{1.96}{v}$ .

Also, $IBL = mg = 1.96$

Hence, $\dfrac{1.96}{v} = 1.96 \text{ } \Rightarrow v = 1m/s$

I did the same way.

Ronak Agarwal - 6 years, 9 months ago

Same here.

Keshav Tiwari - 6 years, 7 months ago

Keshav Kumar
Mar 13, 2015

E=BLV i=BLV (1/R1+1/R2) force acting=iLB=B^2 L^2 V(1/R1+1/R2) IN static condition F=mg R1=(BLV)^2/.76 R2=(BLV)^2/1.2 putting these values we got V(1.96)=mg V=1m/s

Terminal velocity

The answer is 1.

3 solutions

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