# Electromagnetic induction and conducting rods

1. Dec 25, 2004

### apchemstudent

Two 0.68 m long conducting rods are rotating at the same speed in opposite directions, and both are perpendicular to a 4.7 T magnetic field. As the drawing shows, the ends of these rods come to within 1.0mm of each other as they rotate. More-over, the fixed ends about which the rods are rotating are connected by a wire, so these ends are at the same electric potential. If a potential difference of 4.5*10^3 V is required to cause a 1.0 mm spark in air, what is the angular speed (in rad/s) of the rods when a spark jumps across the gap?

Is it possible to still use the formula

Voltage potential = velocity * magnetic field * length of rod ?
v = r*angular speed

(4.5*10^3)/(4.7*.68) = v = r * angular speed

i got the angular speed as 2070 rad/sec. Is this correct how i solved the question? if not can you correct me. Thanks

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2. Dec 25, 2004

### Staff: Mentor

Remember that the base of each rotating rod is at the same potential, but they move in opposite directions so that one will contribute V/2 and the other -V/2, and the potential difference is V = 4500 V.

3. Dec 26, 2004

### apchemstudent

The answer in the back said 2100 rad/sec. At first i thought the same way with V/2, but i only got 1035 rad/sec. So doubling it will bring the answer closer. However though, is this the proper way to solve the problem, as posted from above?

4. Dec 26, 2004

### Andrew Mason

Yes, but the speed is a function of l, so you have to apply a bit of calculus:

$$dE = vBdl \rightarrow E = \int_0^L \omega lBdl = \frac{1}{2}\omega BL^2$$

Note: the other rod is rotating with speed $-\omega$ so the potential from the centre to the end is $- \frac{1}{2}\omega BL^2$

The condition for spark is Potential Difference = 4.5kV: $E_L - E_R = 4,500$.

AM

Last edited: Dec 26, 2004
5. Dec 26, 2004

### Gamma

Also note that one can avoid integration by dealing only with the angular velocity as it is independant of the length.

The angle swept by the rod in one second = $$\omega$$

So the area swept by the rod in one second, $$A=\frac{1}{2}L^2\omega$$

$$E = -\frac {d\phi} {dt} = -B\frac {dA} {dt}$$

$$E = -\frac {1} {2}BL^2\omega$$

Regards,
Gamma.