Magnetic Force: Class 12 Physics Notes, Definition, Working Principle, Formula & Real-Life Applications

6. Magnetic Force on A Current Carrying Wire 

Suppose a conducting wire, carrying a current i, is placed in a magnetic field $\vec{B}$ . Consider a small element dl of the wire (figure). The free electrons drift with a speed v_d opposite to the direction of the current. The relation between the current i and the drift speed v_d is

i = jA = nev_dA.        ....(i)

Here A is the area of cross-section of the wire and n is the number of free electrons per unit volume. Each electron experiences an average (why average?) magnetic force

The number of free electrons in the small element considered in nAdl. Thus, the magnetic force on the wire of length dl is

$$\begin{array}{l}\overrightarrow{\mathrm{dF}}=\left(nAd\ell \right)(-e{\vec{v}}_d\times \vec{B})\\ \text{If we denote the length}d\ell \text{along the direction of the current by}\overrightarrow{d\ell }\text{, the above equation becomes:}\\ \overrightarrow{\mathrm{dF}}=nAe{\vec{v}}_{d}d\ell \times \vec{B}.\\ \text{Using (j),}\overrightarrow{\mathrm{dF}}=i\overrightarrow{d\ell }\times \vec{B}.\\ \text{The quantity}i\overrightarrow{d\ell }\text{is called a current element.}\\ {\vec{F}}_{\text{res}}=\int \overrightarrow{\mathrm{dF}}=\int i\overrightarrow{d\ell }\times \vec{B}=i\int \overrightarrow{d\ell }\times \vec{B}\text{(∵ i is same at all points of the wire.)}\\ \text{If}\vec{B}\text{is uniform then}{\vec{F}}_{\text{res}}=i(\int \overrightarrow{d\ell })\times \vec{B}\\ {\vec{F}}_{\text{res}}=i\vec{L}\times \vec{B}\\ \text{Here}\vec{L}=\int \overrightarrow{d\ell }=\text{vector length of the wire = vector connecting the end points of the wire.}\end{array}$$

$$\begin{array}{l}\text{If a current loop of any shape is placed in a uniform}\vec{B}\text{then}{\overrightarrow{F_{\text{res}}}}_{\text{magnetic}}\text{on it}=0\\ \text{(∴}\vec{L}=0\text{)}\end{array}$$