Consider two different metallic strips (1 and 2) of same dimensions (length , width and thickness ) with carrier densities and , respectively. Strip 1 is placed in magnetic field and strip 2 is placed in magnetic field , both along positive -directions

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 $\vec{B}$ 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 $P Q R S$ and appearance of equal and opposite charges on the face opposite to $P Q R S$ . 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.

Q. 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 option(s) is(are) [2015]

Smart Achievers · Accepted Answer

(1, 3)

$∵$ $V = \frac{I}{n e d} \times B$

$∴$ $V \propto \frac{B}{n}$ $\Rightarrow \frac{V_{1} n_{1}}{B_{1}} = \frac{V_{2} n_{2}}{B_{2}}$

If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ $\Rightarrow V_{2} = 2 V_{1}$

and if $M$ and $n_{1} = n_{2}$ $\Rightarrow V_{2} = 0.5 V_{1}$

Solution

1 If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ , then $V_{2} = 2 V_{1}$

2 If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ , then $V_{2} = V_{1}$

3 If $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ , then $V_{2} = 0.5 V_{1}$

4 If $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ , then $V_{2} = V_{1}$

Ans.
(1, 3)

$∵$ $V = \frac{I}{n e d} \times B$

$∴$ $V \propto \frac{B}{n}$ $\Rightarrow \frac{V_{1} n_{1}}{B_{1}} = \frac{V_{2} n_{2}}{B_{2}}$

If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ $\Rightarrow V_{2} = 2 V_{1}$

and if $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ $\Rightarrow V_{2} = 0.5 V_{1}$

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Solution

1 If B1=B2 and n1=2n2, then V2=2V1

2 If B1=B2 and n1=2n2, then V2=V1

3 If B1=2B2 and n1=n2, then V2=0.5V1

4 If B1=2B2 and n1=n2, then V2=V1