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JEE Advance - Physics (2015 - Paper 2 Offline - No. 19)

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 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.

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 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.

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 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 statements is/are
If w1 = w2 and d1 = 2d2, then V2 = 2V1
If w1 = w2 and d1 = 2d2, then V2 = V1
If w1 = 2w2 and d1 = d2, then V2 = 2V1
If w1 = 2w2 and d1 = d2, then V2 = V1

Explanation

At equilibrium,

$$ \begin{aligned} & \mathrm{F}_{\mathrm{M}} =\mathrm{F}_{\mathrm{E}} \\\\ & q \mathrm{v} \mathrm{B} =q \mathrm{E} \\\\ & \mathrm{E} =\mathrm{v} \cdot \mathrm{B} \\\\ & i =n e \mathrm{~A} v_d \end{aligned} $$

where $v_d=$ drift speed of electrons and $n=$ electron density

$$ \begin{aligned} \mathrm{E} & =\frac{i}{n e \mathrm{~A}} \mathrm{~B} \\ & =\frac{i}{n e(\omega d)} \end{aligned} $$

Area of cross section is always taken normal to the direction of flow of current. Potential difference,

$\mathrm{V}=\mathrm{E} \omega=\frac{i \mathrm{~B}} { ned }$

For the two strips

$ \mathrm{V}_1=\frac{i \mathrm{~B}}{n e}\left(\frac{1}{d_1}\right)$

$$\mathrm{V}_2=\frac{i B}{n e}\left(\frac{1}{d_2}\right)$$

For option (A)

$$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=\frac{1}{2} $$

For option (B)

$$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=1, \frac{d_2}{d_1}=\frac{1}{2} $$

For option (C)

$$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{2}, \frac{d_2}{d_1}=1 $$

For option (D)

$$ \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=1 $$

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