UY1: Magnetic Force On A Current Carrying Conductor



magnetic force on current carrying conductor

Consider a straight segment of a conducting wire, with length $l$ and cross-sectional area $A$. The wire is in a uniform magnetic field $\vec{B}$, which is perpendicular to the plane of the diagram and directed into the plane.

Average magnetic force on each charge is given by:

$$\begin{aligned} \vec{F} &= q \vec{v}_{d} \times \vec{B} \\ F &= qv_{d}B \end{aligned}$$

Total magnetic force on all the moving charges is given by: (n is the no. of charge carriers per unit volume)

$$\begin{aligned} F &= \text{Total no. of electrons} \times qv_{d}B \\ &= nAl \times qv_{d}B \\ &= nqv_{d}A \times lB \end{aligned}$$

Recall that: $J = \frac{I}{A} = nqv_{d}$ which gives $I = nqv_{d}A$.

Hence,

$$F = IlB$$

 

If the $\vec{B}$ field is not perpendicular to the wire but makes an angle $\phi$ with it,

$$\begin{aligned} F &= IlB_{\perp} \\ &= IlB \sin{\phi} \end{aligned}$$

The general form will be:

$$\vec{F} = I \vec{l} \times \vec{B}$$

Magnetic force on an infinitesimal wire section will be:

$$d\vec{F} = I \, d\vec{l} \times \vec{B}$$

 

Next: Magnetic Force On A Curved Conductor

Previous: Hall Effect

Back To Electromagnetism (UY1)

Back To University Year 1 Physics Notes



Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.



Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.