Verified Solution • 2025–26 Session moderate

QJEEAdv15Phb3ea: JEE Advanced 2015 Physics - Class 12 JEEAdv15Phb3ea Solution: Consider two different metallic strips (1 and 2) of the same…

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

Multiple Choice Options

  • A If w1 = w2 and d1 = 2d2, then V2 = 2V1
  • B If w1 = w2 and d1 = 2d2, then V2 = V1
  • C If w1 = 2w2 and d1 = d2, then V2 = 2V1
  • D If w1 = 2w2 and d1 = d2, then V2 = V1

Step-by-Step Verified Solution

At equilibrium,

amp;FM=FEamp;qvB=qEamp;E=v⋅Bamp;i=ne Avd\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 vd=v_d= drift speed of electrons and n=n= electron density

Eamp;=ine A Bamp;=ine(ωd)\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,

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

For the two strips

V1=i Bne(1d1)\mathrm{V}_1=\frac{i \mathrm{~B}}{n e}\left(\frac{1}{d_1}\right)

V2=iBne(1d2)\mathrm{V}_2=\frac{i B}{n e}\left(\frac{1}{d_2}\right)

For option (A)

V1 V2=d2d1=12\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=\frac{1}{2}

For option (B)

V1 V2=1,d2d1=12\frac{\mathrm{V}_1}{\mathrm{~V}_2}=1, \frac{d_2}{d_1}=\frac{1}{2}

For option (C)

V1 V2=12,d2d1=1\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{2}, \frac{d_2}{d_1}=1

For option (D)

V1 V2=d2d1=1\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{d_2}{d_1}=1

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