In a thin rectangular metallic strip a constant current I flows along the positive $$x-direction$$, as shown in the figure. The length, width and thickness of the strip are l, w and d respectively.A uniform magnetic field B→ is applied on the strip along the positive $$y-direction$$. Due to this, the charge carriers experience a net deflection along the $$z-direction$$. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the $$z-direction$$ is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Consider two different metallic strips (1$$ and $$2) of the same material. Their lengths are the same, widths are $$w1$$ and $$w2$$ and thickness are $$d1$$ and $$d2$$, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the $$x-y$$ plane (see$$ $$figure). $$V1$$ and $$V2$$ are the potential differences between K and M in strips $$1$$ and $$2$$, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct $$statement(s)$$ is (are)Consider$$ two different metallic strips $$(1$$ and $$2) of same dimensions (length$$ l, width w and thickness $$d) with carrier densities $$n1$$ and $$n2$$, respectively. Strip $$1$$ is placed in magnetic field $$B1$$ and strip $$2$$ is placed in magnetic field $$B2$$ , both along positive $$y-directions$$. Then $$V1$$ and $$V2$$ are the potential differences developed between K and M in strips $$1$$ and $$2$$, respectively. Assuming that the current I is the same for both the strips, the correct $$option(s)$$ is (are)

Question

In a thin rectangular metallic strip a constant current I flows along the positive  $$x-direction$$, as shown in the figure.  The length, width and thickness of the strip are  l, w and d respectively.A uniform magnetic field  B→ is applied on the strip along the positive  $$y-direction$$.  Due to this, the charge carriers experience a net deflection along the  $$z-direction$$.  This results in accumulation of charge carriers on the surface  PQRS and appearance of equal and opposite charges on the face opposite to  PQRS.  A potential difference along the  $$z-direction$$ is thus developed.  Charge accumulation continues until the magnetic force is balanced by the electric force.  The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Consider two different metallic strips (1$$ and $$2) of the same material.  Their lengths are the same, widths are $$w1$$  and  $$w2$$ and thickness are $$d1$$  and  $$d2$$, respectively.  Two points K  and M are symmetrically located on the opposite faces parallel to the $$x-y$$  plane (see$$ $$figure). $$V1$$   and  $$V2$$ are the potential differences between  K and M  in strips $$1$$ and $$2$$, respectively.  Then, for a given current I flowing through them in a given magnetic field strength B, the correct $$statement(s)$$ is (are)Consider$$ two different metallic strips $$(1$$ and $$2) of same dimensions (length$$  l, width  w and thickness $$d) with carrier densities $$n1$$  and  $$n2$$, respectively.  Strip $$1$$ is placed in magnetic field  $$B1$$ and strip $$2$$ is placed in magnetic field $$B2$$ , both along positive  $$y-directions$$.  Then  $$V1$$ and  $$V2$$ are the potential differences developed between  K and  M in strips $$1$$ and $$2$$, respectively.  Assuming that the current  I is the same for both the strips, the correct $$option(s)$$ is (are)

Infinity Learn · Accepted Answer

$$FB=Bev=BeinAe=BnAFe=eENow$$ , $$Fe=FBeE=BnA$$⇒$$E=BnAePotential$$ difference, $$V=Ed$$⇒$$V=BnAe$$⋅$$w=Bwn(wd)e=BnedV1V2=d2d1$$⇒ if $$w1=w2$$ and $$d1=2d2then$$ $$2V1=V2V1V2=d2d1$$⇒ if $$w1=2w2$$ and $$d1=d2then$$ $$V1=V2V=BnedV1V2=B1B2$$×$$n2n1IfB1=B2$$ and $$n1=2n2$$ then $$V2=2V1IfB1=2B2$$ and $$n1=n2$$ then $$V2=0.5V1$$

Motion of a charged particle

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$\begin{array}{l} V = \frac{B}{n e d} \\ \frac{V_{1}}{V_{2}} = \frac{B_{1}}{B_{2}} \times \frac{n_{2}}{n_{1}} \\ I f B_{1} = B_{2} and n_{1} = 2 n_{2} then V_{2} = 2 V_{1} \\ I f B_{1} = 2 B_{2} and n_{1} = n_{2} then V_{2} = 0.5 V_{1} \end{array}$

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Motion of a charged particle

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FB=Bev=BeinAe=BnAFe=eENow , Fe=FBeE=BnA⇒E=BnAePotential difference, V=Ed⇒V=BnAe⋅w=Bwn(wd)e=BnedV1V2=d2d1⇒ if w1=w2 and d1=2d2then 2V1=V2V1V2=d2d1⇒ if w1=2w2 and d1=d2then V1=V2

V=BnedV1V2=B1B2×n2n1IfB1=B2 and n1=2n2 then V2=2V1IfB1=2B2 and n1=n2 then V2=0.5V1

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$\begin{array}{l} V = \frac{B}{n e d} \\ \frac{V_{1}}{V_{2}} = \frac{B_{1}}{B_{2}} \times \frac{n_{2}}{n_{1}} \\ I f B_{1} = B_{2} and n_{1} = 2 n_{2} then V_{2} = 2 V_{1} \\ I f B_{1} = 2 B_{2} and n_{1} = n_{2} then V_{2} = 0.5 V_{1} \end{array}$