Mechanics or what? - 3

Electricity and Magnetism Level 4

A conducting rod of mass m m and length L L is free to slide on smooth horizontal conducting rails as shown. Resistance of the resistor is R. Uniform magnetic field intensity B B exists through out the region and is normally inward as shown. The rod is given a initial velocity v o v_o . Find the velocity of the rod as a function of time.

inspiration

Try my previous problem here

v = v o e − B 2 L 2 t 3 m R 2 \large v=v_oe^{\frac{-B^2L^2t}{3mR^2}} v = v o e − B 2 L 2 t R 2 \large v=v_oe^{\frac{-B^2L^2t}{R^2}} v = e − B 2 L 2 t m R \large v=e^{\frac{-B^2L^2t}{mR}} v = v o e − B 2 L t m R + v o \large v=v_oe^{\frac{-B^2Lt}{mR}}+v_o v = v o e − B 2 L 2 t m R \large v=v_oe^{\frac{-B^2L^2t}{mR}} None of these choices v = v o e − B 2 L 2 t 4 m R \large v=v_oe^{\frac{-B^2L^2t}{4mR}} v = v o e B 2 L 2 t m R \large v=v_oe^{\frac{B^2L^2t}{mR}}

Sparsh Sarode by Sparsh Sarode , 22, India

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2 solutions

Rishabh Jain
Jun 30, 2016

F = − B i l = − B ( B v l ) l R = − B 2 v l 2 R F=-Bil=-\dfrac{B(Bvl)l}R=-\dfrac{B^2vl^2}{R} ⟹ a = F m = − B 2 v l 2 m r = d v d t \implies a=\dfrac Fm=-\dfrac{B^2vl^2}{mr}=\dfrac{dv}{dt}

Separating variables and putting limits: ∫ v 0 v d v v = − B 2 l 2 m r ∫ 0 t t d t \int_{v_0}^v \dfrac{dv}{v}=\dfrac{-B^2l^2}{mr}\int_0^t tdt ⟹ ln ⁡ ( v v 0 ) = − B 2 l 2 t m R \implies \ln\left(\dfrac{v}{v_0}\right)=\dfrac{-B^2l^2t}{mR} v = v o e − B 2 L 2 t m R \boxed{\large v=v_oe^{\frac{-B^2L^2t}{mR}}}

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Sorry, but same solution... Every problem I try to solve, you would have solved it by that time

Vignesh S - 4 years, 11 months ago

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This has been posted long back... :-) . Its cool that most of times in many problems our thinking process is similar. :-P

Rishabh Jain - 4 years, 11 months ago

Yeah!! That's true

Vignesh S - 4 years, 11 months ago
Ayon Ghosh
Jan 25, 2018

Alternatively one does it by using energy conservation ; Loss in K.E. of rod appears as power in Resistor.

− d K d t = P r e s i s t o r = V 2 R = B 2 v 2 L 2 R = − m v d v d t \dfrac{-dK}{dt} = P_{resistor} = \dfrac{V^2}{R} = \dfrac{B^2v^2L^2}{R} = -mv \dfrac{dv}{dt}

d v d t = − B 2 L 2 v m R ; ∫ v 0 v d v v = ∫ 0 t − B 2 L 2 d t m R \dfrac{dv}{dt} = \dfrac{-B^2L^2v}{mR} ; \displaystyle\int_{v_{0}}^v \dfrac{dv}{v} = \displaystyle\int_0^t \dfrac{-B^2L^2dt}{mR}

v = v o e − B 2 L 2 t m R \large v=v_oe^{\frac{-B^2L^2t}{mR}}

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