Amit & Deepali Sinha Foundation

Weighing the Carriers by Braking

Irodov 3.206DifficultyUniqueness

Problem Irodov 3.206

A coil of radius r=25r = 25 cm, wound from a thin copper wire of length l=500l = 500 m, spins about its own axis with angular velocity ω=300\omega = 300 rad/s. Its ends are joined to a ballistic galvanometer through sliding contacts, and the whole circuit has resistance R=21R = 21 Ω\Omega. When the coil is suddenly stopped, a charge q=10q = 10 nC passes through the galvanometer. Find the specific charge (charge to mass ratio) of the current carriers in copper.

The coil face-on, radius r, turning at ω about its axis through O, and the ballistic galvanometer G, whose terminals meet the leads from the coil's two ends at the sliding contacts (the paired open terminals).

The coil face-on, radius rr, turning at ω\omega about its axis through OO, and the ballistic galvanometer GG, whose terminals meet the leads from the coil’s two ends at the sliding contacts (the paired open terminals).

Given
  • r=25r = 25 cm
  • l=500l = 500 m
  • ω=300\omega = 300 rad/s
  • R=21 ΩR = 21~\Omega
  • q=10q = 10 nC
Find
the specific charge e/me/m of the carriers

Solution

Step 1Braking the wire drives the carriers

Let each carrier have charge ee and mass mm. While the coil turns, the wire moves along its own length with speed v=ωrv = \omega r, and the free carriers move with it. When the coil is braked, the lattice is stopped by the brake, but the carriers are free: to slow down with the wire each needs a force m dv/dtm\,dv/dt along the wire. In the frame of the wire this is an inertial force on each carrier, acting just as an extraneous field would:

E∗=−me dvdt(1) E^* = -\frac{m}{e}\,\frac{dv}{dt} \tag{1}
Step 2The emf and the current

The field (1) is the same at every point of the wire and points along it, so round the whole length ll it gives an emf, and the current through the circuit of resistance RR follows:

E=E∗ l=−me l dvdtI=ER=−m le R dvdt\begin{aligned} \mathcal{E} &= E^*\,l = -\frac{m}{e}\,l\,\frac{dv}{dt}\\ I &= \frac{\mathcal{E}}{R} = -\frac{m\,l}{e\,R}\,\frac{dv}{dt} \end{aligned}
Step 3The charge through the galvanometer

The ballistic galvanometer measures the time integral of the current. Integrating while the speed falls from ωr\omega r to zero:

q=∫I dt=−m le R∫ωr0dv=me l ωrRq = \int I\,dt = -\frac{m\,l}{e\,R}\int_{\omega r}^{0} dv = \frac{m}{e}\,\frac{l\,\omega r}{R}

Solving this for e/me/m and putting in the numbers:

em=(500 m)(300 s−1)(0.25 m)(10×10−9 C)(21 Ω)\frac{e}{m} = \frac{(500~\text{m})(300~\text{s}^{-1})(0.25~\text{m})}{(10\times10^{-9}~\text{C})(21~\Omega)}
em=l ωrqR≈1.8×1011 C/kg\frac{e}{m} = \frac{l\,\omega r}{qR} \approx 1.8\times10^{11}~\text{C/kg}
(2)

This is the specific charge of the electron, so the carriers in copper are electrons.

Check

Dimensions: l ωrl\,\omega r is m2^2/s, and qRqR is C Ω\,\Omega = V \,s, so the ratio is m2^2/(V s2^2); a volt is a joule per coulomb, kg m2^2/(s2^2 C), so this is C/kg. Independence of the braking: only the total change of speed ωr\omega r entered, so however the brake acts it gives the same qq; the stop need only be quick against the galvanometer’s period.

Remark

This is the Tolman and Stewart experiment of 1916. The effect is tiny: by (1) a deceleration of 11 m/s2^2 makes an effective field of only about 6×10−126\times10^{-12} V/m, which is why the experiment needs a long wire, a high speed and a sensitive galvanometer. The sign of qq tells the sign of the carriers: the charge flows the way a negative carrier running on ahead of the wire would carry it.