Question

A metal cube with sides of length a is moving at velocity vec{v}=v_0\hat{y} across a uniform magnetic field \vec{B}=B_0\hat{z}. The cube is oriented so that four of its edges areparallel to its direction of motion (i.e., the normal vector of twofaces are parallel to the direction of motion). View Figure



Find E_vec, the electric field inside the cube.

Now, instead of electrons, suppose that the free charges havepositive charge q.Examples include "holes" in semiconductors and positive ions inliquids, each of which act as "conductors" for their freecharges.

If one replaces the conducting cube with one that has positivecharge carriers, in what direction does the induced electric fieldpoint?

Answer 1

Concepts and reason

The concept required to solve the given problem is Lorentz force.

First, calculate the magnetic force acting on the charged particle. Then determine the induced electric field by using the concept of Lorentz force. After that, determine the direction of induced electric field due to positive charge by using right hand thumb rule.

Fundamentals

The magnetic force is given by,

${\vec F_M} = q\left( {\vec v \times \vec B} \right)$

Here, $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field.

The electric force acting on a charge particle placed in an electric field is given as,

${\vec F_E} = q\vec E$

Here, E is the electric field.

The Lorentz force is given by,

${\vec F_L} = {\vec F_E} + {\vec F_M}$

Right hand Thumb Rule: The direction of the cross product may be found by right hand thumb rule. It states that if the index finger points in the direction of the velocity vector and the middle finger points in the direction of magnetic field vector, then the thumb will point in the direction of the force vector.

(1)

The figure given below shows the cube moving through the magnetic field.

Here, $a$ is the length of the side of the cube, ${v_0}$ is the velocity with which the block moves through the region of magnetic field ${B_0}$ .

The magnetic force acting on the charged particle is given by,

${\vec F_M} = q\left( {\vec v \times \vec B} \right)$

Here, $q$ is the charge, $v$ is the velocity, $\theta$ is the angle between velocity and magnetic field vector and $B$ is the magnetic field.

The cube moves with a velocity $\vec v$ perpendicular to the magnetic field $\vec B$ which forces the electrons to move towards the face ABCD. Due to the electrons moving to face ABCD a deficiency of negative charges gets created on face PQRS due to which positive charge develops there.

Due to the negative charge developed on the face ABCD and positive charge on the face PQRS, the electric field will be directed into the page of paper.

Since the electron is moving with a velocity perpendicular to the magnetic field. Hence, the angle between the velocity vector and magnetic field vector will be $90^\circ$ .

Substitute ${v_0}\hat y$ for $\vec v$ and ${B_0}\hat z$ for $\vec B$ in the equation ${\vec F_M} = q\left( {\vec v \times \vec B} \right)$ .

$\begin{array}{c}\\{{\vec F}_M} = q\left( {{v_0}\hat y \times {B_0}\hat z} \right)\\\\ = q{v_0}{B_0}\hat x\\\end{array}$

The Lorentz force is given by,

${\vec F_L} = {\vec F_E} + {\vec F_M}$

In steady state the net force on the charges will be zero. Hence, substitute 0 for ${\vec F_L}$ in the above equation and rearrange the equation for ${\vec F_E}$ .

$\begin{array}{c}\\0 = {{\vec F}_E} + {{\vec F}_M}\\\\{{\vec F}_E} = - {{\vec F}_M}\\\end{array}$

Substitute $q{v_0}{B_0}\hat x$ for ${\vec F_M}$ and $q\vec E$ for ${\vec F_E}$ in the above equation and rearrange the equation for $\vec E$ .

$\begin{array}{c}\\q\vec E = - \left( {q{v_0}{B_0}\hat x} \right)\\\\\vec E = - {v_0}{B_0}\hat x\\\end{array}$

(2)

The positive charge is moving in the positive x direction with velocity v and the magnetic field is in positive z direction. So, from right hand thumb rule the magnetic force will be in positive x direction. For the particle to be in steady state, the electric force on the particle should be in the direction opposite to the direction of magnetic force. Thus, the direction of magnetic force will be negative x.

As for a positive charge electric field is in the direction of electric force, so the direction of electric field will be negative x direction.

Hence, the direction of induced electric field will be negative x direction i.e. $- \hat x$ .

Ans: Part 1

The induced electric field will be $- {v_0}{B_0}\hat x$ .

Part 2

The direction of induced electric field will be negative x direction i.e. $- \hat x$ .