derive an expression for magnetic force on a current carrying conductor placed in a uniform magnetic field

Dear Student,

Please find below the solution to the asked query:

By Lorenz force equation, if a charged particle of charge "q" is moving in magnetic field of strength "B" with a velocity "V", then the force acts on it is,

$\overrightarrow{F_B} = q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$
cIf instead of a charge if we consider a current carrying conductor as shown in the figure,


Consider a small element of wire as shown. Then the amount of charge can be written by,
$dq = idt$
So, force on this element is,
$$\begin{gathered} d\overrightarrow{F_B} = i dt\left(\overrightarrow{V}\times \overrightarrow{B}\right) \Rightarrow d\overrightarrow{F_B} = i \left(\left(\overrightarrow{V}dt\right)\times \overrightarrow{B}\right) \Rightarrow d\overrightarrow{F_B} = i \left(\overrightarrow{dl}\times \overrightarrow{B}\right) \\ If \overrightarrow{B} = cons\tan t \\ \Rightarrow \overrightarrow{F_B} = \int d\overrightarrow{F_B} = i \left(\int \overrightarrow{dl}\times \overrightarrow{B}\right) \\ \Rightarrow \overrightarrow{F_B} = i\left(\overrightarrow{L}\times \overrightarrow{B}\right) \end{gathered}$$

If the conductor is a straight wire of length "l", then force on it is,
 

$F_B = iBl \sin \theta$
Where, θ = angle between the current direction in the wire and the direction of magnetic field.

Direction of force is decided based on the Fleming's Left hand Rule.

Hope this information will clear your doubts about the topic.

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