Describe the motion of a charged particle moving in a uniform magnetic field that is perpendicular to its velocity, and use this description to derive and apply the magnetic‑force formula for a straight current‑carrying conductor.
\[
B=\frac{F}{IL\sin\theta}\qquad\left[\text{unit: tesla (T)}\right]
\]
where \(F\) is the magnetic force on a straight wire of length \(L\) carrying current \(I\), and \(\theta\) is the angle between the wire (current direction) and the field.
The magnetic force on a charge \(q\) moving with velocity \(\mathbf v\) in a magnetic field \(\mathbf B\) is
\[
\mathbf F = q\,\mathbf v \times \mathbf B
\]
When the velocity is perpendicular to the magnetic field (\(\phi = 90^{\circ}\)) the magnetic force is always at right angles to the motion and therefore acts as a centripetal force.
\[
qvB = \frac{mv^{2}}{r}\qquad\Longrightarrow\qquad
r = \frac{mv}{qB}
\]
\[
\omega = \frac{v}{r}= \frac{qB}{m}
\]
Note that \(\omega\) (and the linear frequency \(f=\omega/2\pi\)) depends only on the charge‑to‑mass ratio and the magnetic field – it is independent of the speed.
In a conductor the charge carriers (usually electrons) drift with speed \(v_{d}\). For a wire of length \(L\), cross‑sectional area \(A\) and carrier number density \(n\):
\[
I = nqAv_{d}\qquad\text{(definition of current in terms of drift speed)}
\]
The total magnetic force on all carriers in the segment is
\[
F{\text{total}} = (nqAv{d})\,L\,B\sin\theta = I L B\sin\theta
\]
This leads directly to the general magnetic‑force formula for a straight conductor.
\[
\boxed{F = B I L \sin\theta}
\]

Place the left hand so that the first finger points along \(\mathbf B\) and the second finger along the direction of current \(I\); the thumb then points in the direction of the force \(\mathbf F\) on the wire.
A current‑carrying straight wire generates a circular magnetic field around it. The magnitude at a distance \(r\) from the wire is
\[
B = \frac{\mu{0} I}{2\pi r}\qquad\left(\mu{0}=4\pi\times10^{-7}\,\text{T·m·A}^{-1}\right)
\]
| Quantity | Symbol | Expression | Notes |
|---|---|---|---|
| Magnetic flux density | \(B\) | \(B=\dfrac{F}{IL\sin\theta}\) | unit = tesla (T) |
| Magnetic force on a charge | \(\mathbf F\) | \(q\,\mathbf v\times\mathbf B\) | right‑hand rule for \(q>0\) |
| Force magnitude ( \(v\perp B\) ) | \(F\) | \(qvB\) | |
| Radius of circular path | \(r\) | \(\displaystyle\frac{mv}{qB}\) | proportional to momentum |
| Cyclotron (angular) frequency | \(\omega\) | \(\displaystyle\frac{qB}{m}\) | independent of speed |
| Current (drift speed) | \(I\) | \(nqAv_{d}\) | \(n\)= carrier density |
| Force on a straight conductor | \(F\) | \(BIL\sin\theta\) | maximal at \(\theta=90^{\circ}\) |
| Magnetic field around a straight wire | \(B\) | \(\displaystyle\frac{\mu_{0}I}{2\pi r}\) | circular field lines |
Radius of a proton’s path
\(m = 1.67\times10^{-27}\,\text{kg}\), \(q = +e = 1.60\times10^{-19}\,\text{C}\), \(v = 2.0\times10^{6}\,\text{m s}^{-1}\), \(B = 0.5\,\text{T}\) (perpendicular).
\[
r = \frac{mv}{qB}
= \frac{(1.67\times10^{-27})(2.0\times10^{6})}{(1.60\times10^{-19})(0.5)}
\approx 4.2\times10^{-2}\,\text{m}
\]
Force on a wire at an angle
\(L = 0.30\;\text{m}\), \(I = 5.0\;\text{A}\), \(B = 0.5\;\text{T}\), \(\theta = 30^{\circ}\).
\[
F = B I L \sin\theta
= (0.5)(5.0)(0.30)\sin30^{\circ}
= 0.375\;\text{N}
\]
Measuring \(B\) with a known force
\(L = 0.40\;\text{m}\), \(I = 2.0\;\text{A}\), measured force \(F = 0.80\;\text{N}\) (wire ⟂ \(\mathbf B\)).
\[
B = \frac{F}{IL} = \frac{0.80}{(2.0)(0.40)} = 1.0\;\text{T}
\]



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