Question

A charged particle having charge q experiences a force $\vec{F} = q( - \hat{j} +\hat{k} )$ N in a magnetic field B when it has a velocity $v_1 = 1 \hat{i}$ m/s. The force becomes $\vec{F} = q( \hat{i} - \hat{k} )$ N when the velocity is changed to $v_2 = 1 \hat{j}$ m/s. The magnetic induction vector at that point is :
A. $( \hat{i} + \hat{j} + \hat{k} )$ T
B. $( \hat{i} - \hat{j} - \hat{k} )$T
C. $( - \hat{i} - \hat{j} + \hat{k} )$ T
D. $( \hat{i} + \hat{j} - \hat{k} )$T

Answer 1

We need to find Magnetic induction which is denoted by B. It will have components along x, y and z directions. The cross product of velocity vector and magnetic field vector will give a resultant Lorentz force. Break the cross product into three component equations.

Formula used:
The Lorentz force acting on a charge particle moving in a magnetic field is given as:
$\vec{F} = q \vec{v} \times \vec{B}$

Complete answer:
For a charged particle q, which is entering a magnetic field $\vec{B}$ with a velocity $\vec{v}$, the force acting on the particle is given as:
$\vec{F} = q \vec{v} \times \vec{B}$
When we break it into component form we can write it as:
$F_x \hat{i} + F_y \hat{j} + F_z \hat{k} = q \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ v_x & v_y & v_z\\\ B_x & B_y & B_z \end{vmatrix}$
Now substituting the given values for two cases:
(1) From the question, the components for the first case can be written as:
$F_x = 0, F_y = -q, F_z = q$ ;
$v_x = 1, v_y = 0, v_z = 0$.
Which gives us:
$0 \hat{i} -q \hat{j} + q \hat{k} = q \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ 1 & 0 & 0\\\ B_x & B_y & B_z \end{vmatrix}$

Cancelling q on both sides and solving the cross product gives us:
$-1 \hat{j} + 1 \hat{k} = -B_z \hat{j} + B_y \hat{k}$
Equating the components we directly get:
$B_y = 1$
$B_z = 1$
Now, similarly proceeding to the second case.
(2) $F_x = q, F_y = 0, F_z = -q,$ ;
$v_x = 0, v_y = 1, v_z = 0,$
Which gives us:
$q \hat{i} + 0 \hat{j} + -q \hat{k} = q \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\\ 0 & 1 & 0\\\ B_x & B_y & B_z \end{vmatrix}$
After simplification we get:
$1 \hat{i} - 1 \hat{k} = B_z \hat{i} - B_x \hat{k}$
Equating the components we directly get:
$B_x = 1$
$B_z = 1$
Therefore, we can also check here that in case (1) and case (2) we get the same value for $B_z$ component. So upon putting together all the obtained components, we get:
$\vec{B} = 1 \hat{i} + 1 \hat{j} + 1 \hat{k}$T.

Therefore, the correct answer is option (A).

Note:
The notation for magnetic fields is often taken to be $\vec{B}$ which is also magnetic induction. It is the magnetic field produced within a substance due to some free and bound currents in it. Whereas another notation $\vec{H}$ represents magnetic fields produced within a substance due to free currents. In the hysteresis loop too, it is the $\vec{H}$ that we are able to control for a substance.