Flemings Left Hand Rule Class 12 states that if we stretch the thumb, the forefinger and the middle finger of our left hand such that they are mutually perpendicular to each other.
If the forefinger gives the direction of current and middle finger points in the direction of magnetic field then the thumb points towards the direction of the force or motion of the conductor.Then if the fore-finger points in the direction of field $(\overrightarrow{\mathrm{B}})$, the central finger in the direction of current , the thumb will point in the direction of force.(ii) Right-hand Palm Rule : Stretch the fingers and thumb of right hand at right angles to each other. Then if the fingers point in the direction of field $\overrightarrow{\mathrm{B}}$ and thumb in the direction of current $\mathbf{I}$, the normal to palm will point in the direction of force.Regarding the force on a current-carrying conductor in a magnetic field it is worth mentioning that :(a) As the force BI d\ell sin \theta is not a function of position r, the magnetic force on a current element is non-central [a central force is of the form $\mathrm{F}=\mathrm{Kf}(\mathrm{r}) \overrightarrow{\mathrm{n}_{\mathrm{r}}}]$(b) The force d\vec $\overrightarrow{\mathrm{F}}$ is always perpendicular to both $\overrightarrow{\mathrm{B}}$ and $\mathrm{Id} \vec{\ell}$ though $\overrightarrow{\mathrm{B}}$ and $\mathrm{Id} \vec{\ell}$ may or may not be perpendicular to each other.(c) In case of current-carrying conductor in a magnetic field if the field is uniform i.e.,$\overrightarrow{\mathrm{B}}=$ constt.,$\overrightarrow{\mathrm{F}}=\int \mathrm{I} \mathrm{d} \vec{\ell} \times \overrightarrow{\mathrm{B}}=\mathrm{I}\left[\int \mathrm{d} \vec{\ell}\right] \times \overrightarrow{\mathrm{B}}$and as for a conductor $\int_{\mathrm{d}} \vec{\ell}$ represents the vector sum of all the length elements from initial to final point, which in accordance with the law of vector addition is equal to the length vector $\overrightarrow{\ell^{\prime}}$ joining initial to final point, so a current-carrying conductor of any arbitrary shape in a uniform field experience a force$\vec{F}=I\left[\int d \vec{\ell}\right] \times \vec{B}=I \ell^{\prime} \times \vec{B}$where $\vec{\ell}$ is the length vector joining initial and final points of the conductor as shown in fig.(d) If the current-carrying conductor in the form of a loop of any arbitrary shape is placed in a uniform field,$\overrightarrow{\mathrm{F}}=\oint \mathrm{Id\vec{\ell }} \times \overrightarrow{\mathrm{B}}=\mathrm{I}[\oint \overrightarrow{\mathrm{d} \vec{\ell}}] \times \overrightarrow{\mathrm{B}}$and as for a closed loop, $\oint \mathrm{d} \vec{\ell} \text { is always zero. [vector sum of all } \mathrm{d} \vec{\ell}]$i.e., the net magnetic force on a current loop in a uniform magnetic field is always zero as shown in fig.Here it must be kept in mind that in this situation different parts of the loop may experience elemental force due to which the loop may be under tension or may experience a torque as shown in fig.

Current Loop in a Uniform Field

(e) if a current-carrying conductor is situated in a non-uniform field, its different elements will experience different forces; so in this situation,
$\overrightarrow{\mathrm{F}_{\mathrm{R}}} \neq 0$ but $\overrightarrow{\mathrm{\tau}}$ may or may not be zero
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