Physics · Ch 3 — Magnetism and Magnetic Effects of Electric Current

Force on a Current Carrying Conductor Placed in a Magnetic Field

3.10.5

Force on a Current Carrying Conductor Placed in a Magnetic Field

Inside a current-carrying conductor of cross-section $A$ , free electrons drift (opposite to the conventional current direction) with drift speed $v_d$ , related to the current by $I = neAv_d$ (Unit 2), where $n$ is the free-electron density. In an external field $\vec B$ , each drifting electron feels a Lorentz force; summing this over all $N=nAdl$ electrons in a small element of length $dl$ gives the force on that element,

d\vec F = I\,d\vec l \times \vec B

Integrating over the whole conductor (length $l$ , field assumed uniform) gives the total force on a straight current-carrying wire,

$\boxed{\vec F = I\,\vec l \times \vec B}, \qquad\text{in magnitude } F = BIl\sin\theta$ …

Figure 3.51Current-carrying conductor in a magnetic field

What this figure shows. A straight conducting wire of cross-sectional area A and length L is drawn carrying current I through a field B pointing into the page; a small elemental segment of length dl is marked on the wire, with a force arrow FB drawn on it, representing the net Lorentz force summed over the very large number of drifting free electrons inside that short segme …

Figure 3.52Fleming's Left Hand Rule

What this figure shows. A left hand is drawn with the thumb, forefinger and middle finger held mutually perpendicular to each other. The forefinger is labelled 'Field' (pointing along B), the middle finger is labelled 'Current' (pointing along I), and the thumb is labelled 'Force' (giving the direction of the force the current-carryin …

Misc Example 3.24Current to hold a rod stationary on an incline

Worked out. A metallic rod of linear mass density 0.25 kg/m rests on a frictionless 45-degree incline, held from sliding by a current I flowing through it in a vertical field of 0.25 T. Resolving forces along the incline, equilibrium requires mg sin(45) = BIl cos(45), so I = (m/l) g tan(45)/B = (0.25)(9.8)(1)/(0.25), which gives I=9.8 A -- the current needed so that the magnetic force's component along the slope exactly balances the rod's weight component pulling …

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