Looking for classes? Ksquare Career Institute, Bengaluru →
A very long straight wire carries a current I. At the instant when a charge $+Q$ at point P has velocity $\vec{V}$, as shown, the force on the charge is -
[add image]
[add image]
A
Along ox
B
Opposite to oy
C
Along oy
D
Opposite to ox
Detailed Solution
The magnetic force on a moving charge is $\vec{F} = Q(\vec{v} \times \vec{B})$.
In the figure the velocity of the charge is along the x-axis ($\hat{i}$), and at P the magnetic field of the straight current-carrying wire (given by the right hand thumb rule) points into the plane of the paper, i.e. along $-\hat{k}$.
$\vec{F} = QvB(\hat{i} \times (-\hat{k}))$
Since $\hat{i} \times \hat{k} = -\hat{j}$, we get $\hat{i} \times (-\hat{k}) = +\hat{j}$.
$\vec{F} = QvB\,\hat{j}$
So the force on the charge is along OY.
In the figure the velocity of the charge is along the x-axis ($\hat{i}$), and at P the magnetic field of the straight current-carrying wire (given by the right hand thumb rule) points into the plane of the paper, i.e. along $-\hat{k}$.
$\vec{F} = QvB(\hat{i} \times (-\hat{k}))$
Since $\hat{i} \times \hat{k} = -\hat{j}$, we get $\hat{i} \times (-\hat{k}) = +\hat{j}$.
$\vec{F} = QvB\,\hat{j}$
So the force on the charge is along OY.