A conducting strip of width 1.5 mm is in a magnetic field.As a result, there is a potential difference of 4.3 mV across the width of the strip.What is the magnitude of the electric field in the strip?
A)0.35 V/m
B)1.2 V/m
C)1.9 V/m
D)2.9 V/m
E)6.4 V/m

Question

Quizplus · Accepted Answer

We can use the formula for the Hall voltage: $V_H=Bvd$, where $B$ is the magnetic field, $v$ is the velocity of the charge carriers, and $d$ is the width of the strip. Rearranging this formula to solve for the electric field, we get $E=\frac{V_H}{d}\cdot\frac{1}{B}$.

Plugging in the values given, we get $E=\frac{4.3	ext{ mV}}{1.5	ext{ mm}}\cdot\frac{1}{B}$. We don't know the value of $B$, but we can find it using the fact that the Hall voltage is caused by the Lorentz force $F_B=qvB$, where $q$ is the charge of the carriers. If we assume that the charge carriers are electrons, then $q=-1.6	imes10^{-19}	ext{ C}$.

We also know that the current density $J$ is related to the carrier density $n$, the electron charge $q$, and the carrier velocity $v$ by the formula $J=nqv$. Rearranging this formula, we can solve for $v=\frac{J}{nq}$, and then use this expression in the formula for the Lorentz force: $F_B=q\frac{J}{nq}B=\frac{JB}{n}$.

The Hall voltage is equal to the electric field times the width of the strip times the charge carrier density $n$: $V_H=E\cdot d\cdot nq$. Rearranging this formula, we can solve for $n=\frac{V_H}{Eqd}$.

Now we can combine all of these formulas to solve for $B$ and then $E$. We have:

$\frac{JB}{n}=-F_B=-\frac{V_H}{d}=-\frac{4.3	ext{ mV}}{1.5	ext{ mm}}$
$n=\frac{J}{qv}=\frac{J}{qn^{-1}\frac{J}{nq}}=n$
$V_H=E\cdot d\cdot nq=\frac{E\cdot d}{B}\cdot\frac{JB}{q}\cdot q$

Plugging in the given values of $d$, $V_H$, and $q$, and using the known value of the current density $J=I/A$, where $I$ is the current and $A$ is the cross-sectional area of the strip, we get:

$n=\frac{I}{qA}\cdot\frac{1}{E}$

$\frac{JB}{n}=-\frac{V_H}{d}$ becomes:

$\frac{B}{n}=-\frac{V_H}{dJ}\cdot\frac{1}{q}=-\frac{4.3	ext{ mV}}{1.5	ext{ mm}}\cdot\frac{1}{1.5	ext{ mm}\cdot I/A}\cdot\frac{1}{1.6	imes10^{-19}	ext{ C}}=-\frac{1.07}{IA}$

$\frac{B}{n}=-\frac{1.07}{IA}$ can be substituted into $n=\frac{I}{qA}\cdot\frac{1}{E}$ to get:

$n=-\frac{qA}{1.07I}\cdot E$

Now we can solve for $E$:

$E=\frac{V_H}{d}\cdot\frac{1}{B}=\frac{4.3	ext{ mV}}{1.5	ext{ mm}}\cdot\frac{1}{-\frac{1.07}{nqIA}}=2.9	ext{ V/m}$

Therefore, the magnitude of the electric field in the strip is 2.9 V/m. The correct answer is D).

A Conducting Strip of Width 1