Question

Q12

I have Q in Phyiscs 2 and i need soluation with steps

A small disk with radius R is made of conducting material. The disk is immersed in a magnetic field B and has the ability to spin about an axis parallel to B with angular speed omega Assuming the induces EMF acts along radial lines (originates from the center of the disk to its edge) and B is to the plane of the disk, show the induced EMF is given by: V = EMF = Br2omega / 2 If you wanted to power a small 1.5 volt calculator, determine the maximum rotational speed omega (in rpm) thew disk would have to spin at if radius were 12 cm and the magnetic field present is 40 mu t.

Answer 1

Part a)

Variuous elements of the system described in the quaestion are illustrated in the figure above.

B the magnetic field perdicular to the plane of the conductor disc.

$\omega$ angular velocity of the disc rotation (anticlockwise, as shown)

r the radius of the disc

Now, the electrons in the conductor (disc) feels a linear velocity $v= r\omega$ , and its direction is as shown in the figure.

Moreover, the these electron also feels a force (also called Lorentz force) due to this linear velocity given by

$F = q(v\times B) = q(r\omega B)$

where q is the charge of an electron and its direction is along the radial direction outward from the center. (NB: this direction could be inward too, if B is directed downward). Here the direction of the force may be tricky as charge of electron negative.

Due to this Lorentz force, there arises an effective eletric field (=force per unit charge) of magnitude:

$E = r\omega B$

Now, this effective electric filed gives rise to an induced emf (potential difference) beween the origin and rim of the disc, given by

$V = - \int_{0}^{r} E dr = -B\omega \int_{0}^{r} r dr = -\frac{B\omega r^2}{2}$

Here we have used the relation $E = -\frac{dV}{dr}$ and the -ve sign in the expression of Induced emf V only signifies the direction of the potential difference (say in this set up disc rim becomes center become +ve and with center being -ve) again here this direction of induced emf may be changed by changing the direction of rotation of the disc and the magnetic field.

Hence the magnitude of the indiced emf is

$V = \frac{B\omega r^2}{2}$

Part b)

$B = 40\mu T = 40\times 10^{-6}T$

hence

$\omega = \frac{2V}{Br^2} = \frac{2 \times 1.5}{40\times 10^{-6}\times 0.12^2}=5.20\times 10^6 \rm rad/s$

or

$\omega = 49.66 \times 10^6 \rm rpm$