Measuring properties of some material An experiment was carried out to determine

Question

Measuring properties of some material An experiment was carried out to determine the electrical properties of a new conducting material. A bar was made out of the material, L = 20 cm long with a rectangular cross section h = 4 cm high and w = 0.9 cm deep. The bar was part of a circuit and carried a steady current (the diagram shows only part of the circuit). A uniform magnetic field of 1.3 tesla was applied perpendicular to the bar, coming out of the page (using some coils that are not shown). Two voltmeters were connected along and across the bar as shown, The reading on Voltmeter 1 is DeltaV_1 = minus0.45 volt. The reading on Voltmeter 2 is Delta V_2 = +0.00028 volt. The connections across the bar for Voltmeter 2 were carefully placed directly across from each other to eliminate false readings corresponding to the much larger voltage along the bar, There is only one kind of mobile charge in this material. Remember that a voltmeter gives a positive reading if the positive lead is connected to a higher potential location than the negative lead. What is the sign of the mobile charges, and which way do they move? What is the drift speed v of the mobile charges? m/s What is the mobility u of the mobile charges? (m/s)/(V/m) The current running through the bar was measured to be 0.8 ampere. If each mobile charge is singly charged (|q| = e), how many mobile charges are there in 1 m^3, of this material? M^minus3 What is the resistance in ohms of this length of bar? Ohm

Explanation / Answer

For this problem, we have to take into account the Hall Effect, to solve it.

A) For the voltmeter 1, which reading is negative (-0.45 V), it implies that extreme at the left of the bar are at a electric potential more high than the right extreme. So, the charges are negative and them will move from the right to the left.

For voltmeter 2, which reading is positive (+0.00028 V), it implies that extreme at top are at more high potential than extreme down. So, in this way, charges will move from down to the top.

B) The drift speed v of the mobile charges is given by the Hall's electric field, which formula is:

VH = vBL , where v is velocity and L = lenght of the bar. Clearing v from this equation: v =  VH / BL =>

v = 0.45 V / (1.5 T)(20x10-2 m) => v = 1.73 x 106m/s

for extreme which positive volyage, we apply the same reasoning: v = 0.00028 V / (1.5 T)(4 x 10-2 m) =>

v = 4.6 x 103 m/s

D) The quantity of charges mobile will be give by the next equation: Vxy = iB / nqw => n = iB / Vxy qw

n = 0.8 A (1.5 T) / 0.45V (1.6 x10-19 C)(0.9 x 10-2 m) =

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