Mechanics or what? - 4

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 ' x x '.

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None v = v o e B 2 L 2 x m R v=v_oe^{\frac{B^2L^2x}{mR}} v = v o e − B 2 L 2 x m R v=v_oe^{\frac{-B^2L^2x}{mR}} v = v o + B 2 L 2 x m R v=v_o+\dfrac{B^2L^2x}{mR} v = v o − B 2 L 2 x m R v=v_o-\dfrac{B^2L^2x}{mR}

Sparsh Sarode by Sparsh Sarode , 22, India

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1 solution

Rishabh Jain
Jul 1, 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 v̸ l 2 m r = − v̸ d v d x \implies a=\dfrac Fm=-\dfrac{B^2\not vl^2}{mr}=-\not v\dfrac{dv}{dx}

Separating variables and integrating with appropriate limits: ∫ v 0 v d v = − B 2 l 2 m R ∫ 0 x d x \int_{v_0}^vdv=\dfrac{-B^2l^2}{mR}\int_0^xdx ⟹ v − v 0 = − B 2 l 2 x m R \implies v-v_0= -\dfrac{B^2l^2x}{mR}

⟹ v = v 0 − B 2 l 2 x m R \implies \boxed{v=v_0-\dfrac{B^2l^2x}{mR}}

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