Animated Solution for Physics - Magnetic Effects of Current: Comprehension Passage

Electrons in a Magnetic Dance

Imagine a rectangular metallic strip acting as a highway for electric current. The current $I$ flows smoothly to the right, along the positive $x$-axis. But remember, in metals, the actual charge carriers are electrons. Because they carry a negative charge, a current to the right means the electrons are actually drifting to the left, along the negative $x$-axis, with a drift velocity ${\mathbf{v}}_d=-v_d\hat{i}$.

Now, let's introduce a twist: a uniform magnetic field $\mathbf{B}$ pointing straight up, along the positive $y$-axis ($\mathbf{B}=B\hat{j}$). As these electrons drift through the magnetic field, they experience a Lorentz force.

Using the right-hand rule, the cross product of their velocity and the magnetic field ($-\hat{i}\times \hat{j}$) gives a direction of $-\hat{k}$. However, because the electron's charge is negative, the final magnetic force ${\mathbf{F}}_m$ points in the positive $z$-direction:

Driven by this magnetic force, the electrons are pushed towards the front face of the strip (the surface $PQRS$ at $z=w$). As they accumulate there, the front face becomes negatively charged, leaving a net positive charge on the back face ($z=0$).

This separation of charges instantly creates an electric field $\mathbf{E}$ pointing from the positive back face to the negative front face (along the positive $z$-axis). This electric field fights back, exerting an electric force ${\mathbf{F}}_e=-e\mathbf{E}$ on the electrons in the negative $z$-direction.

Very quickly, a state of dynamic equilibrium is reached where the electric force perfectly balances the magnetic force:

The potential difference, known as the Hall voltage $V$, developed across the width $w$ of the strip is simply the electric field multiplied by the distance:

To make this equation truly useful, we need to express it in terms of the macroscopic current $I$. We know that current is related to drift velocity by $I=neA{v}_d$, where $n$ is the carrier density and $A$ is the cross-sectional area.

Since the current flows along the $x$-axis, it cuts through an area defined by the width $w$ and the thickness $d$. So, $A=w\times d$. Rearranging for drift velocity gives $v_d=\frac{I}{newd}$.

Substituting this back into our voltage equation yields a beautiful cancellation:

Notice a fascinating physical insight here: the Hall voltage is completely independent of the width $w$ of the strip! It only depends on the thickness $d$ parallel to the magnetic field.

Armed with our master equation, $V=\frac{IB}{ned}$, the questions become a breeze.

For Question 14: We are comparing strips of the same material, meaning the carrier density $n$ is constant. With $I$ and $B$ also constant, the voltage is inversely proportional to the thickness $d$ ($V\propto \frac{1}{d}$). - If we halve the thickness ($d_1=2d_2$), the voltage doubles ($V_2=2V_1$). This makes option (a) correct. - If we change the width but keep the thickness the same ($d_1=d_2$), the voltage remains unchanged ($V_2=V_1$). This makes option (d) correct.

For Question 15: Now the dimensions are identical, so $d$ is constant. The voltage is directly proportional to the magnetic field $B$ and inversely proportional to the carrier density $n$ ($V\propto \frac{B}{n}$). - If we halve the carrier density ($n_1=2n_2$), the voltage doubles ($V_2=2V_1$). This makes option (a) correct. - If we halve the magnetic field ($B_1=2B_2$), the voltage halves ($V_2=0.5V_1$). This makes option (c) correct.