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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 the accumulation of charge carries 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 untill 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 , width 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 statements is/are

A) if

$w_{1}=w_{2}$ and

$d_{1}=2d_{2}$ , then

$V_{2}=2V_{1}$

B) If

$w_{1}=w_{2}$ and

$d_{1}=2d_{2}$ , then

$V_{2}=V_{1}$

C) If

$w_{1}=2w_{2}$ and

$d_{1}=d_{2}$ , then

$V_{2}=2V_{1}$

D) If

$w_{1}=2w_{2}$ and

$d_{1}=d_{2}$ , then

$V_{2}=V_{1}$

Answer:

Option A,D

Explanation:

$F_{B}=B_{ev}=B_{e}\frac{1}{nAe}=\frac{BI}{nA}$

Fe= eE

$\Rightarrow$ F_e =F_B

$eE=\frac{BI}{nA}$

$\Rightarrow$ $E=\frac{B}{nAe}$

$V=Ed=\frac{BI}{nAe}.w=\frac{BIw}{n(wd)e}=\frac{BI}{ned}$

$\frac{V_{1}}{V_{2}}=\frac{d_{2}}{d_{1}}$

$\Rightarrow$ $E=\frac{B}{nAe}$

$V=Ed=\frac{BI}{nAe}.w=\frac{BIw}{n(wd)e}=\frac{BI}{ned}$

$\frac{V_{1}}{V_{2}}=\frac{d_{2}}{d_{1}}$

$\Rightarrow$ if w₁ =2w₂ and d₁ =d₂

V₁ =V₂

$\therefore$ Correct answes are (a) and (d)

Discussion Forum

Answer:

Explanation: