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Lenz's Law

In a thin rec...

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 $\to B$ 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 same dimensions (length $l$ , width $w$ and thickness $d$ ) with carrier densities $n_{1}$ and $n_{2}$ , respectively. Strip 1 is placed in magnetic field $B_{1}$ and strip~2 is placed in magnetic field $B_{2}$ , both along positive $y$ -directions. Then $V_{1}$ and $V_{2}$ 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 options(s) is(are),

B_{1} = B_{2}

and

n_{1} = 2 n_{2}

, then

V_{2} = 2 V_{1}

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B_{1} = B_{2}

and

n_{1} = 2 n_{2}

, then

V_{2} = V_{1}

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B_{1} = 2 B_{2}

and

n_{1} = n_{2}

, then

V_{2} = 0.5 V_{1}

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B_{1} = 2 B_{2}

and

n_{1} = n_{2}

, then

V_{2} = V_{1}

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Solution

The correct option is C If $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ , then $V_{2} = 0.5 V_{1}$ .
$V_{m} - V_{k} = \frac{I B}{n e d}$

The ratio of potential difference in two strips is given by, $\begin{matrix} \frac{V_{2}}{V_{1}} = \frac{B_{2}}{B_{1}} \frac{n_{1}}{n_{2}} . \end{matrix}$ Substitute $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ in equation to get $V_{2} = 2 V_{1}$ .

Substitute $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ in above equation to get $V_{2} = 0.5 V_{1}$ .

Thus, A and C are correct answers.

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