Force On A Conductor Carrying Current In A Magnetic Field - Practice Questions & MCQ

Magnetic force on a current carrying conductor - 

In case of current carrying conductor in a magnetic field force experienced by its small length element is $d \vec{F}=i(d \vec{l} \times \vec{B})$

                                                                     
For total force, we will integrate the above equation. So the total magnetic force -

$
\vec{F}=\int d \vec{F}=\int i(d \vec{l} \times \vec{B})
$

If magnetic field is uniform i.e., $\vec{B}=$ constant and $\int d \vec{l}=\vec{L}=$ vector sum of all the length elements from initial to final point. Which is in accordance with the law of vector addition is equal to length vector $\bar{L}^{\prime}$ joining initial to final point.

Then,

$
\vec{F}=i\left[\int d l\right] \times \vec{B}=i(\vec{L} \times \vec{B})
$
 

Direction of force - 

The direction of force is perpendicular to both the length and magnetic field vector as we have discussed earlier that the result of the cross product of two vector have direction perpendicular to both the vectors. It can be find by right hand palm rule, screw rule, right hand thumb rule etc. Here we will discuss one important rule for this i.e., Fleming’s left-hand rule.

According to Fleming’s left-hand rule - Stretch the fore-finger, central finger and thumb left hand mutually perpendicular. Then if the fore-finger points in the direction of field $\vec{B}$  and the central in the direction of current i, the thumb will point in the direction of force. For better understanding, look at the image given below, 

Note - If curved wire is given in the question then the length will be taken as shown in the figure - 

And the direction of the length vector should be in the direction of current.