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Indian Institute of Technology Kharagpur

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Problem 63

Show that when a metal rod $L$ meters long moves at speed $v$ perpendicular to $\vec{B}$ field lines, the magnetic force exerted by the field on the electrically charged particles in the rod produces a potential difference between the ends of the rod equal to the product $B I y$ .

Answer

$\Delta V=B v L$

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## Discussion

## Video Transcript

In this particular case, you have to consider a system like this where you have Ah, Rod, This way. This is your rod. Now you have magnetic field in the upward direction and your rod is traveling. We have a lasted TV in let's say that direction. Okay, so now the electric field induced at the end of the rod is given Bari Genuine sucks over change in time. In this particular because there is only one soup. So I am just going to write sox ease, be a Because the area is perpendicular to the magnetic field. Does the tea now BCE constant and the area of the rod is given by They said This is X and let's say this direction is worry. So that's Delta off X y over desert E. It was be why can come out, Because why is constant You have got their X over dlt. Actually, that's not used the variable Why it is given that the rent of the rod is l. The length of the rod is l so that's my l. And it is not changing, so l can come out of the parent disease and I have dexterity which is the velocity off the Do not. So this is how you can show that motion of the IMF produces and a mound off BFV electric field across the to end off the road.

## Recommended Questions

Prove that the velocity of charged particles moving along a straight path through perpendicular electric and magnetic fields is $v=E / B$ . Thus crossed electric and magnetic fields can be used as a velocity selector independent of the charge and mass of the particle involved.

Integrated Concepts

Prove that the velocity of charged particles moving along a straight path through perpendicular electric and magnetic fields is $v=E / B$. Thus crossed electric and magnetic fields can be used as a velocity selector independent of the charge and mass of the particle involved.

The rod shown in the accompanying figure is moving through a uniform magnetic field of strength $B=0.50 \mathrm{T}$ with a constant velocity of magnitude $v=8.0 \mathrm{m} / \mathrm{s} .$ What is the potential difference between the ends of the rod? Which end of the rod is at a higher potential?

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$$

|\mathcal{E}|=\frac{\mu_{0} I v}{2 \pi} \ln \left(1+\frac{\ell}{r}\right)

$$

A conducting rod of length $\ell$ moves with velocity $\overrightarrow{\mathbf{v}}$ parallel to a long wire carrying a steady current $I$ . The axis of the rod is maintained perpendicular to the wire with the near end a distance $r$ away (Fig. P31.65). Show that the magnitude of the emf induced in the rod is

$$|\boldsymbol{\varepsilon}|=\frac{\mu_{0} I v}{2 \pi} \ln \left(1+\frac{\ell}{r}\right)$$

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A thin rod of length $\ell$ and uniform charge per unit length $\lambda$ lies along the $x$ axis as shown in Figure $\mathrm{P} 23.37$ . (a) Show that the electric field at $P,$ a distance $d$ from the rod along its perpendicular bisector, has no $x$ component and is given by $E=2 k_{e} \lambda \sin \theta_{0} / d .$ (b) What If? Using your result to part (a), show that the field of a rod of infinite length is $E=$ 2$k_{e} \lambda / d$ .

A 25-cm rod moves at 5.0 m/s in a plane perpendicular to a magnetic field of strength 0.25 T. The rod, velocity vector, and magnetic field vector are mutually perpendicular, as indicated in the accompanying figure. Calculate (a) the magnetic force on an electron in the rod, (b) the electric field in the rod, and (c) the potential difference between the ends of the rod. (d) What is the speed of the rod if the potential difference is $1.0 \mathrm{V}$ ?

Use Newtonian circular motion concepts to show that the radius $r$ of the circle in which a charged particle spirals while moving perpendicular to a magnetic field is proportional to the particle's speed $y$ .

A particle carrying a 50 -\muC charge moves with velocity $\vec{v}=5.0 \hat{\imath}+3.2 \hat{k} \mathrm{m} / \mathrm{s}$ through a magnetic field given by $\vec{B}=9.4 \hat{\imath}+6.7 \hat{\jmath}$ T. (a) Find the magnetic force on the particle. (b) Form the dot products $\vec{F} \cdot \vec{v}$ and $\vec{F} \cdot \vec{B}$ to show explicitly that the force is perpendicular to both $\vec{v}$ and $\vec{B}$.