JEE Challenger
More from Moving Charges and Magnetism
PhyMedAdv 2015 • Paper 2

Hall Potential Difference Dependence on Metallic Strip Dimensions

Comprehension Passage

In a thin rectangular metallic conductor strip, a steady current II flows along the positive xx-axis, as shown in the illustration. The length, width, and thickness of this strip are denoted by ll, ww, and dd, respectively. A uniform external magnetic field B⃗\vec{B} acts along the positive yy-direction. Consequently, the mobile charge carriers experience a transverse magnetic deflection along the zz-direction. This causes accumulation of charge carriers on the planar surface PQRSPQRS and induces an equal and opposite charge layer on the surface opposite to PQRSPQRS. Therefore, a potential difference develops along the zz-direction. The accumulation of charges persists until the transverse magnetic force is counterbalanced by the electrostatic force. Assume that the electric current is distributed uniformly across the strip's cross-section and is conducted solely by electrons.

Passage Diagram 1

Consider two metallic strips (1 and 2) made of identical material. Their lengths are equal, with widths w1w_1 and w2w_2, and thicknesses d1d_1 and d2d_2, respectively. Two probe points KK and MM are located symmetrically on opposing faces parallel to the xx-yy plane (refer to the figure). Let V1V_1 and V2V_2 be the potential differences established between KK and MM in strips 1 and 2, respectively. For a fixed current II flowing through each strip under a specified magnetic field strength BB, which of the following statement(s) is/are correct?

Options

A

If w1=w2w_1 = w_2 and d1=2d2d_1 = 2d_2, then V2=2V1V_2 = 2V_1

Correct
B

If w1=w2w_1 = w_2 and d1=2d2d_1 = 2d_2, then V2=V1V_2 = V_1

C

If w1=2w2w_1 = 2w_2 and d1=d2d_1 = d_2, then V2=2V1V_2 = 2V_1

D

If w1=2w2w_1 = 2w_2 and d1=d2d_1 = d_2, then V2=V1V_2 = V_1

Correct

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceElectric Field

Step-by-Step Solution

To find the potential difference VV developed between points KK and MM, let us analyze the forces acting on the charge carriers (electrons) in the strip:

  1. Current and Drift Velocity: The current II flowing along the positive xx-axis through the strip of cross-sectional area A=w⋅dA = w \cdot d is given by: I=neAvd=newdvdI = n e A v_d = n e w d v_d where:

    • nn is the charge carrier density (number of electrons per unit volume),
    • ee is the elementary charge,
    • vdv_d is the drift velocity of the electrons along the negative xx-axis.

    Solving for the drift velocity: vd=Inewdv_d = \frac{I}{n e w d}

  2. Equilibrium Condition (Hall Effect): The electrons experience a magnetic Lorentz force in the zz-direction due to the magnetic field B⃗=Bj^\vec{B} = B\hat{j}: F⃗B=−e(v⃗d×B⃗)=−e(−vdi^×Bj^)=evdBk^\vec{F}_B = -e (\vec{v}_d \times \vec{B}) = -e \left(-v_d \hat{i} \times B \hat{j}\right) = e v_d B \hat{k}

    This transverse deflection accumulates negative charges on one face and positive charges on the opposite face, setting up an opposing transverse electric field E⃗H=−Ezk^\vec{E}_H = -E_z \hat{k}.

    At equilibrium, the electric force balances the magnetic force: eEz=evdB  ⟹  Ez=vdBe E_z = e v_d B \implies E_z = v_d B

  3. Potential Difference: The probe points KK and MM lie on opposite faces parallel to the xx-yy plane, separated by distance ww along the zz-axis. Therefore, the Hall potential difference VV between KK and MM is: V=Ez⋅w=vdBwV = E_z \cdot w = v_d B w

    Substituting the expression for vdv_d: V=(Inewd)Bw=IBnedV = \left(\frac{I}{n e w d}\right) B w = \frac{I B}{n e d}

From this result, we see that for strips of the same material (identical nn) carrying the same current II in the same magnetic field BB:

  • VV is independent of the width ww.
  • VV is inversely proportional to the thickness dd, i.e., V∝1dV \propto \frac{1}{d}.

Thus, the ratio of potential differences is: V2V1=d1d2\frac{V_2}{V_1} = \frac{d_1}{d_2}

Evaluating the given cases:

  • For w1=w2w_1 = w_2 and d1=2d2d_1 = 2d_2: V2V1=2d2d2=2  ⟹  V2=2V1(Option A is correct)\frac{V_2}{V_1} = \frac{2d_2}{d_2} = 2 \implies V_2 = 2V_1 \quad \text{(Option A is correct)}
  • For w1=2w2w_1 = 2w_2 and d1=d2d_1 = d_2: V2V1=d2d2=1  ⟹  V2=V1(Option D is correct)\frac{V_2}{V_1} = \frac{d_2}{d_2} = 1 \implies V_2 = V_1 \quad \text{(Option D is correct)}

Correct Options: A, D

PreviousNext