Question

The figure shows cross-sectional view of an infinite cylindrical wire with a cavity in it, having uniform current density, $\overrightarrow{j}=-j_0\hat{k}$.

• A. Magnetic field inside the cavity is uniform.
• B. Magnetic field inside the cavity is along $\overrightarrow{a}$.
• C. Magnetic field inside the cavity is perpendicular to $\overrightarrow{a}$.
• D. If an electron is projected with velocity $v_0\hat{j}$ inside the cavity, it will move undeviated.

Answer 1

Answer: (A), (C), (D)

If an electron is projected with velocity $v_0\hat{j}$ inside the cavity, it will move undeviated.

The magnetic field inside the cavity in the infinitely long cylindrical conductor is given by,

$\overrightarrow{B}=\frac{{\mu }_0(\overrightarrow{j}\times \overrightarrow{a})}{2}$

Where, $\overrightarrow{a}=$ separation between the axis of cylinder and the axis of the cavity.

As $j$ and $a$ are constant, the magnetic field inside the cavity is uniform.

Direction is given by $\hat{j}\times \hat{a}.$

Hence, $\overrightarrow{B}$ will be perpendicular to both $\overrightarrow{j}$ and $\overrightarrow{a}$.

Here, $\hat{B}=\hat{j}\times \hat{a}=-\hat{k}\times \hat{i}=-\hat{j}$

If an electron is projected inside the cavity with velocity $v_0\hat{j}$, then,

$F=qvB\sin \theta$

$\therefore F=0[\because \theta ={180}^{\circ }]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, options $\left(A\right),\left(C\right)$ and $\left(D\right)$ are correct.