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  • A conducting bar of mass m and length is given a velocity <img alt="\mathrm{v_0}" src="/latex-image/?%5Cmathrm%7Bv_0%

A conducting bar of mass m and length \ell is given a velocity \mathrm{v_0} along a smooth conducting horizontal rail having a resistance R as shown in the figure. What is the maximum distance traversed by the bar before coming to rest in an inward magnetic field of induction B. ?

Option: 1

\mathrm{\frac{m R v_0}{B^2 r^2}}


Option: 2

\mathrm{\frac{2 m R v_0}{B^2 \ell^2}}


Option: 3

\mathrm{\frac{m R v_0}{2 B^2 \ell^2}}


Option: 4

None


Answers (1)

best_answer

The induced voltage across the bar \mathrm{=\mathrm{E}=\mathrm{B} \ell \mathrm{V}}

The induced current along the bar \mathrm{=i=\frac{E}{R}=\frac{B \ell v}{R}}

This current carrying conductor experiences a magnetic force F which opposes the motion (velocity v) that is given as

\mathrm{\begin{aligned} & \mathrm{mv} \frac{\mathrm{dv}}{\mathrm{dx}}=-\left(\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}\right) \ell \mathrm{B} \\ & \Rightarrow \quad \mathrm{vdv}=-\frac{\mathrm{B}^2 \ell^2 \mathrm{v}}{\mathrm{mR}} \mathrm{dx} \\ & \Rightarrow \quad \int_{\mathrm{v}_0}^0 \mathrm{dv}=-\frac{\mathrm{B}^2 \ell^2}{\mathrm{mR}} \int_0^{\mathrm{s}} \mathrm{dx} \\ & \Rightarrow \quad \mathrm{S}=\frac{\mathrm{mR} \mathrm{v}_0}{\mathrm{~B}^2 \ell^2} \end{aligned}}

Posted by

Kuldeep Maurya

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