⚡ فعل الحقل المغناطيسي في التيار الكهربائي (الوحدة 2 - الكهرباء والمغناطيسية 2)

The torque on a frame with current I in field B: \(\tau=NIAB\sin\theta\), where θ is the angle between the frame’s normal and the field. Torque is maximum when \(\sin\theta=1\), i.e. \(\theta=90°\) (the frame’s plane parallel to the field lines), and zero when \(\sin\theta=0\), i.e. \(\theta=0°\) or \(180°\) (the frame’s plane perpendicular to the field).

Derivation: the magnetic force on a charge moving perpendicular to the field: \(F=qvB\) (from the basic Lorentz force law). Rearranging: \(B=\dfrac{F}{qv}\) when the velocity is perpendicular to the field.

Defining the tesla: it is the field strength that exerts a force of 1 newton on a charge of 1 coulomb moving at 1 m/s perpendicular to it: \(1\;T = 1\;\dfrac{N}{C\cdot m/s}\).

Measurement: it relies on the torque on a current-carrying coil placed in a magnetic field. In the moving-coil galvanometer, the field is radial so the angle between the normal and the field is always 90°, giving \(\tau=NIAB\), balanced by a spring’s torque \(\tau=k\theta\). At equilibrium: \(k\theta=NIAB\Rightarrow I=\dfrac{k}{NAB}\theta\) — a linear relation.

Increasing sensitivity: increase the number of turns N, the field B, or the coil area A; decrease the torsion constant k (a thinner, more flexible suspension wire); a mirror-and-light pointer can also magnify the reading.

1) The Laplace force is perpendicular to both the current and the field (right-hand rule): \(F=BIL=0.1\times40\times0.04=0.16\;N\)

2) Work: \(W=F\times d=0.16\times0.15=0.024\;J\)

3) Tilting the rails by θ (with the field — and hence F — staying horizontal), equilibrium requires the gravity component along the incline \((mg\sin\theta)\) to balance the horizontal force’s component along the incline \((F\cos\theta)\):

\(F\cos\theta = mg\sin\theta \;\Rightarrow\; \tan\theta=\dfrac{F}{mg}=\dfrac{0.16}{0.016\times9.8}\approx1.02\;\Rightarrow\;\theta\approx45.6°\)

Note: the common (incorrect) equation \(\sin\theta=F/(mg)\) gives \(1.02>1\) here, which is mathematically impossible — the correct relation for this geometry (F staying horizontal as the plane tilts) is \(\tan\theta=F/(mg)\).

Magnetic force: \(F=BIL=2\times10^{-3}\times10\times0.04=8\times10^{-4}\;N\)

Its torque about the pivot: \(\tau_{mag}=F\times d=8\times10^{-4}\times0.5=4\times10^{-4}\;N\cdot m\)

Weight: \(w=mg=0.05\times9.8=0.49\;N\)

Weight’s torque (small deflection α, center of mass at the wire’s midpoint): \(\tau_{weight}=mg\times\dfrac{L}{2}\sin\alpha=0.49\times0.3\sin\alpha=0.147\sin\alpha\)

Equating the torques at equilibrium: \(0.147\sin\alpha=4\times10^{-4}\;\Rightarrow\;\sin\alpha\approx0.00272\;\Rightarrow\;\alpha\approx0.156°\)

1) Since the frame’s plane is parallel to the field, the normal makes 90° with the field: \(\tau=NIAB=100\times10\times4\times10^{-4}\times2\times10^{-4}=8\times10^{-5}\;N\cdot m\)

2) After rotating 30° from the parallel position, the normal makes 60° with the field:

\(\tau=NI'AB\sin60°=100\times0.002\times4\times10^{-4}\times2\times10^{-4}\times0.866\approx1.386\times10^{-8}\;N\cdot m\)

\(K=\dfrac{\tau}{\theta(rad)}=\dfrac{1.386\times10^{-8}}{\pi/6}\approx2.65\times10^{-8}\;N\cdot m/rad\)

Total flux through the N-turn frame: \(\Phi=NBA\cos60°=100\times2\times10^{-4}\times4\times10^{-4}\times0.5=4\times10^{-6}\;Wb\)

1) The current is radial from the axle to the rim, the field is horizontal into the page, and the force is tangential (perpendicular to the disk, producing a torque).

2) The force on a length element \(dr\) of the radius: \(dF=BI\,dr\). Integrating along the radius from \(0\) to \(R=0.1\;m\) (current constant along the radius): \(F=BIR=2\times10^{-2}\times5\times0.1=0.01\;N\)

3) The torque of a force element about the axis: \(d\tau=r\,dF=BI\,r\,dr\). Integrating from \(0\) to \(R\): \(\tau=\dfrac{1}{2}BIR^2=0.5\times2\times10^{-2}\times5\times(0.1)^2=5\times10^{-4}\;N\cdot m\)

4) To prevent rotation, hang a mass m at the end of a horizontal radius (moment arm R), balancing its weight’s torque against the magnetic torque: \(mgR=\dfrac{1}{2}BIR^2\Rightarrow m=\dfrac{BIR}{2g}=\dfrac{2\times10^{-2}\times5\times0.1}{2\times9.8}\approx5.1\times10^{-4}\;kg\approx0.51\;g\)

The electric force \(qE\) and magnetic force \(qvB\) oppose each other on this charge. If \(v=\dfrac{E}{B}\), the two forces are equal and opposite, cancelling out, so the body moves in a straight line at constant speed (the principle of a velocity selector). If instead the electric field is zero (\(E=0\)), only the magnetic force remains, and the body moves in a circular path.