Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid1Experiment3.9TheHallEffectandMagneticFieldofaSolenoidThe Hall effect can be used to illustrate the effect of a magnetic field on a movingcharge to investigate various phenomena of electric currents in conductors andespecially semi-conductors. In 1879 E. H. Hall observed that when a current-carryingconductor is placed in a magnetic field of magnitude B that has its directionperpendicular to the current, a voltage difference will appear as a result of themagnetic field. This effect is known as the Hall effect, and this voltage is called theHall voltage. This Hall voltage is proportional to the product of the current andcomponent of normal to the current. The Hall effect is the basis of many practicalapplications and devices such as magnetic field measurements, and position andmotion detectors.ExperimentalObjectives(1) Determine the semiconductor type from the polarity of the Hall voltage, knowingthe orientation of all fields and currents in the experimental arrangement(2) Calculate the carrier concentration and mobility from the magnitude of the Hallvoltage and the known experimental variables.(3) Explain the response of the charge carriers in a material to a magnetic field.(4) Eliminate the side effects which can produce voltages between the Hall probes(5) Predict the Hall voltage which would develop under a given set of Hallexperimentvariables.(6) Measure themagnetic field in a long solenoidExperimentalInstrumentsHall effect set-up (FB510 and FB510A), magnetic field meter of solenoids (FB400)Experimental Principles1.HallEffectThe Hall effect occurs because a charged particle moving in a magnetic field is
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 1 Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid The Hall effect can be used to illustrate the effect of a magnetic field on a moving charge to investigate various phenomena of electric currents in conductors and especially semi-conductors. In 1879 E. H. Hall observed that when a current-carrying conductor is placed in a magnetic field of magnitude B that has its direction perpendicular to the current, a voltage difference will appear as a result of the magnetic field. This effect is known as the Hall effect, and this voltage is called the Hall voltage. This Hall voltage is proportional to the product of the current and component of normal to the current. The Hall effect is the basis of many practical applications and devices such as magnetic field measurements, and position and motion detectors. Experimental Objectives (1) Determine the semiconductor type from the polarity of the Hall voltage, knowing the orientation of all fields and currents in the experimental arrangement. (2) Calculate the carrier concentration and mobility from the magnitude of the Hall voltage and the known experimental variables. (3) Explain the response of the charge carriers in a material to a magnetic field. (4) Eliminate the side effects which can produce voltages between the Hall probes (5) Predict the Hall voltage which would develop under a given set of Hall experiment variables. (6) Measure the magnetic field in a long solenoid. Experimental Instruments Hall effect set-up (FB510 and FB510A), magnetic field meter of solenoids (FB400) Experimental Principles 1. Hall Effect The Hall effect occurs because a charged particle moving in a magnetic field is
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid2subject to the Lorentz force given by:(3.9-1)FLorentz = qv × Bwhere g is a signed quantity representing carrier charge, v is the particle velocityvector, and B is the vector magnetic field. The basic Hall measurement is performedon a semiconductor bar with an electric field applied along its long axis, and magneticfield applied perpendicular to it.2. Hall VoltagesExamine the sample shown in Figure 1. A voltage V is applied, giving rise to a fieldEx=V/L.If the sample is n-type, the majority carrier electrons will move oppositethe applied electric field, right to left (-x). The v × B cross product is in the positivey direction for a B field directed upward in the page.The carrier charge in this case isnegative so the force is actually in the -y direction. This force causes the majoritycarrier electrons to be pushed towards the front edge of the sample.The entire sampleremains neutrally charged as the positive charges of the donor ions are nowuncompensated at the back. There is, however, a gradient of charge increasing fromback to front giving rise to a second electric field perpendicular both to the externallyapplied electric field Ex and the B-field. This new electric field opposes furtheraccumulation of electrons (it could be viewed as rejection by electrons alreadythere)The system reaches equilibrium when the force applied on carriers by the secondelectric field Ey equals and opposes the forced due to the B field.q|Ey/ = qlux·Bzl(3.9-2)Ey gives rise to a voltage which can be measured from the front to the back face ofthe sample. This is called the Hall voltage Vh = Ey b (see Figure 3.9-1 for thedefinition of b). In this n-type sample, the voltage measured from front to back in thesample will be negative. If any of the sign definitions change, however, this sign maychange too.Now consider a p-type sample.Here the majority carrier holes move with the appliedelectric field, left to right in Figure 3.9-1. The force due to the magnetic field, qv × Bis inthe-y direction and oncemore carriers are crowded to thefront face of thesample resulting in an electric field. This time, however, because the carriers arepositive, the Hall voltage measured from front to back on the sample will be positive.Thus, the majority carrier type determines the sign of the Hall voltage
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 2 subject to the Lorentz force given by: F𝐿𝑜𝑟𝑒𝑛𝑡𝑧 = 𝑞𝒗 × 𝑩 (3.9-1) where q is a signed quantity representing carrier charge, v is the particle velocity vector, and B is the vector magnetic field. The basic Hall measurement is performed on a semiconductor bar with an electric field applied along its long axis, and magnetic field applied perpendicular to it. 2. Hall Voltages Examine the sample shown in Figure 1. A voltage 𝑉 is applied, giving rise to a field 𝐸𝑥 = 𝑉/𝐿. If the sample is n-type, the majority carrier electrons will move opposite the applied electric field, right to left (-𝑥). The 𝒗 × 𝑩 cross product is in the positive y direction for a B field directed upward in the page. The carrier charge in this case is negative so the force is actually in the -𝑦 direction. This force causes the majority carrier electrons to be pushed towards the front edge of the sample. The entire sample remains neutrally charged as the positive charges of the donor ions are now uncompensated at the back. There is, however, a gradient of charge increasing from back to front giving rise to a second electric field perpendicular both to the externally applied electric field 𝐸𝑥 and the B-field. This new electric field opposes further accumulation of electrons (it could be viewed as rejection by electrons already there). The system reaches equilibrium when the force applied on carriers by the second electric field 𝐸𝑦 equals and opposes the forced due to the B field. 𝑞|𝐸𝑦| = 𝑞|𝑣𝑥 ∙ 𝐵𝑧 | (3.9-2) 𝐸𝑦 gives rise to a voltage which can be measured from the front to the back face of the sample. This is called the Hall voltage 𝑉𝐻 = 𝐸𝑦 𝑏 (see Figure 3.9-1 for the definition of 𝑏). In this n-type sample, the voltage measured from front to back in the sample will be negative. If any of the sign definitions change, however, this sign may change too. Now consider a p-type sample. Here the majority carrier holes move with the applied electric field, left to right in Figure 3.9-1. The force due to the magnetic field, 𝑞𝒗 × 𝑩 is in the −𝑦 direction and once more carriers are crowded to the front face of the sample resulting in an electric field. This time, however, because the carriers are positive, the Hall voltage measured from front to back on the sample will be positive. Thus, the majority carrier type determines the sign of the Hall voltage
Experiment 3.9The Hall Effect and Magnetic Field of a Solenoid3(VHbB,ConstantCurrentSourceIFigure 3.9-1.Hall voltage circuitThe velocity we have been discussing is the carrier drift velocity and is related to thecurrent by:Ix=qnvxAor(3.9-3)Ix= qpvxAfor n-type or p-type sample respectively. Remember that q is a signed quantity. InEquation (3.9-3), Ix is the current in the x-direction due to the applied electric field,n orp is the carrier concentrationper cm, and A is the cross sectional area of thesample in cm? (width times thickness:b d). The quantity y can easily be solved forand the result substituted in Equation (3.9-2), resulting in)Ix BzEy :(n -type)Iql nA前岛(α-type)(3.9-4)Ey=in which the sign of the charge carrier is now expressed explicitlyIf we substitute the product b-d for A in Equation (3.9-4), where b is the samplewidthanddthesamplethickness,wecanmultiplyEquation(3.9-4)bythewidthtoget the Hall voltage:-Ix BzVH = bEy(n -type)Tql nd+Ix Bz ((3.9-5)VH=bEy=ra(p-type)Since VH, Bz, Ix, d and q are all known (by measurement), it is possible to solve for thecarrier concentration n or p, and determine whether the sample is n-type or p-type
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 3 Figure 3.9-1. Hall voltage circuit The velocity we have been discussing is the carrier drift velocity and is related to the current by: 𝐼𝑥 = 𝑞𝑛𝑣𝑥𝐴 or 𝐼𝑥 = 𝑞𝑝𝑣𝑥𝐴 (3.9-3) for n-type or p-type sample respectively. Remember that q is a signed quantity. In Equation (3.9-3), 𝐼𝑥 is the current in the 𝒙-direction due to the applied electric field, n or p is the carrier concentration per cm3 , and A is the cross sectional area of the sample in cm2 (width times thickness:𝑏 ∙ 𝑑). The quantity v can easily be solved for and the result substituted in Equation (3.9-2), resulting in: 𝐸𝑦 = −𝐼𝑥 |𝑞| 𝐵𝑧 𝑛𝐴 (𝑛 − 𝑡𝑦𝑝𝑒) 𝐸𝑦 = +𝐼𝑥 |𝑞| 𝐵𝑧 𝑝𝐴 (𝑝 − 𝑡𝑦𝑝𝑒) (3.9-4) in which the sign of the charge carrier is now expressed explicitly. If we substitute the product 𝑏 ∙ 𝑑 for A in Equation (3.9-4), where b is the sample width and d the sample thickness, we can multiply Equation (3.9-4) by the width to get the Hall voltage: 𝑉𝐻 = 𝑏𝐸𝑦 = −𝐼𝑥 |𝑞| 𝐵𝑧 𝑛𝑑 (𝑛 − 𝑡𝑦𝑝𝑒) 𝑉𝐻 = 𝑏𝐸𝑦 = +𝐼𝑥 |𝑞| 𝐵𝑧 𝑝𝑑 (𝑝 − 𝑡𝑦𝑝𝑒) (3.9-5) Since VH, Bz, Ix, d and q are all known (by measurement), it is possible to solve for the carrier concentration n or p, and determine whether the sample is n-type or p-type
Experiment 3.9The Hall Effect and Magnetic Field of a Solenoid43.HallCoefficientA useful measurement concept is Hall Coefficient, which is defined as:Ey(3.9-6)RH =JxBzWriting Ry in terms of measurable quantities, we get.RH = VHd(3.9-7)IxBzIf we substitute the value of VH from Equation (3.9-5)into Equation (3.9-7), we cansee:-1RH =lalin (n - type)(p- type)(3.9-8)RH :IqpThe above analysis relies upon the idea that all carriers travel with the drift velocity.4. Carrier MobilityAfter we have determined the carrier concentration and type from Equation (3.9-8)we can use some further experimental information to determine the carriermobilityWe can measure the conductivity from Ohm's law:LLIx(3.9-9)g=A-RbdVAC(3.9-10)= lqlμnn or =lqlμppwhich means that the determination of carrier concentration along with theconductivity provide a measurement of the mobility.a[RHoμnnqloor(3.9-11)=[Rloμp =plql5.Measurement of Magnetic FieldThe magnetic field B can easily be obtained by Equation (3.9-7), resulting in:VHdVHBz 3(3.9-12)IxRHIxKHwhereK is called Hall Sensitivity6.SideEffectsThe voltage appearing between the Hall probes is not generally, the Hall voltage alone
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 4 3. Hall Coefficient A useful measurement concept is Hall Coefficient, which is defined as: 𝑅𝐻 = 𝐸𝑦 𝐽𝑥𝐵𝑍 (3.9-6) Writing 𝑅𝐻 in terms of measurable quantities, we get: 𝑅𝐻 = 𝑉𝐻𝑑 𝐼𝑥𝐵𝑍 (3.9-7) If we substitute the value of VH from Equation (3.9-5) into Equation (3.9-7), we can see: 𝑅𝐻 = −1 |𝑞|𝑛 (𝑛 − 𝑡𝑦𝑝𝑒) 𝑅𝐻 = +1 |𝑞|𝑝 (𝑝 − 𝑡𝑦𝑝𝑒) (3.9-8) The above analysis relies upon the idea that all carriers travel with the drift velocity. 4. Carrier Mobility After we have determined the carrier concentration and type from Equation (3.9-8), we can use some further experimental information to determine the carrier mobility. We can measure the conductivity from Ohm’s law: σ = 𝐿 𝐴∙𝑅 = 𝐿𝐼𝑥 𝑏𝑑𝑉𝐴𝐶 (3.9-9) σ = |𝑞|𝜇𝑛𝑛 𝑜𝑟 σ = |𝑞|𝜇𝑝𝑝 (3.9-10) which means that the determination of carrier concentration along with the conductivity provide a measurement of the mobility. 𝜇𝑛 = σ 𝑛|𝑞| = |𝑅𝐻|σ or 𝜇𝑝 = σ 𝑝|𝑞| = |𝑅𝐻|σ (3.9-11) 5. Measurement of Magnetic Field The magnetic field B can easily be obtained by Equation (3.9-7), resulting in: 𝐵𝑍 = 𝑉𝐻𝑑 𝐼𝑥𝑅𝐻 = 𝑉𝐻 𝐼𝑥𝐾𝐻 (3.9-12) where 𝐾𝐻 is called Hall Sensitivity. 6. Side Effects The voltage appearing between the Hall probes is not generally, the Hall voltage alone
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid5There are other galvanomagnetic and thermomagnetic effects (Nernst effect Vn,Rhighleduc effect Vr and Ettingshausen effect Ve)whichcan producevoltagesbetween the Hall Probes. In addition, IR drop (Vo) due to probe misalignment (zeromagnetic field potential) and thermoelectric voltage due to transverse thermal gradientmay be present. All these except the Ettingshausen effect are eliminated by themethod of averaging four readings. The Ettingshausen effect is negligible in materialsin which a high thermal conductivity is primarily due to lattice conductivity or inwhich the thermoelectric power is small. When the voltage between the Hall probes ismeasured for both directions of current, only the Hall voltage and IR drop reverse.Therefore,the average of these readings eliminates the influence of the other effects.Further, when Hall voltage is measured for both the directions of the magnetic field,the IR drop does not reverse and may therefore be eliminated.We have four voltages in the method of averaging four readings.When +B and +lx are applied, the “Hall electric potential difference" is:(3.9-13)V1 = +VH + Ve + V + VR + VoWhen -B and +lx are applied, the Hall electric potential difference" is:(3.9-14)V2 = -VH - Ve - Vn - VR + VoWhen-B and -lx are applied, the“Hall electric potential difference"is(3.9-15)V3 = +VH + Ve - VN - VR - VoWhen +B and -lx are applied, the“Hall electric potential difference" is(3.9-16)V4=-VH- V+ VN+ VR- VoAsa consequence,Vh ~ Vh + Vg = X-+V-V(3.9-17)47.Magnetic Field of a SolenoidA solenoid is a long helical coil of wire through which a current is run in order tocreate a magnetic field. The magnetic field of the solenoid is the superposition of thefields due to the current through each coil. It is nearly uniform inside the solenoid andclose to zero outside and is similar to the field of a bar magnet having a north pole atone end and a south pole at the other depending upon the direction of current flow
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 5 There are other galvanomagnetic and thermomagnetic effects (Nernst effect VN, Rhighleduc effect VR and Ettingshausen effect VE) which can produce voltages between the Hall Probes. In addition, IR drop (V0) due to probe misalignment (zero magnetic field potential) and thermoelectric voltage due to transverse thermal gradient may be present. All these except the Ettingshausen effect are eliminated by the method of averaging four readings. The Ettingshausen effect is negligible in materials in which a high thermal conductivity is primarily due to lattice conductivity or in which the thermoelectric power is small. When the voltage between the Hall probes is measured for both directions of current, only the Hall voltage and IR drop reverse. Therefore, the average of these readings eliminates the influence of the other effects. Further, when Hall voltage is measured for both the directions of the magnetic field, the IR drop does not reverse and may therefore be eliminated. We have four voltages in the method of averaging four readings. When +B and +Ix are applied, the “Hall electric potential difference” is: 𝑉1 = +𝑉𝐻 + 𝑉𝐸 + 𝑉𝑁 + 𝑉𝑅 + 𝑉0 (3.9-13) When -B and +Ix are applied, the “Hall electric potential difference” is: 𝑉2 = −𝑉𝐻 − 𝑉𝐸 − 𝑉𝑁 − 𝑉𝑅 + 𝑉0 (3.9-14) When -B and -Ix are applied, the “Hall electric potential difference” is: 𝑉3 = +𝑉𝐻 + 𝑉𝐸 − 𝑉𝑁 − 𝑉𝑅 − 𝑉0 (3.9-15) When +B and -Ix are applied, the “Hall electric potential difference” is: 𝑉4 = −𝑉𝐻 − 𝑉𝐸 + 𝑉𝑁 + 𝑉𝑅 − 𝑉0 (3.9-16) As a consequence, 𝑉𝐻 ≈ 𝑉𝐻 + 𝑉𝐸 = 𝑉1−𝑉2+𝑉3−𝑉4 4 (3.9-17) 7. Magnetic Field of a Solenoid A solenoid is a long helical coil of wire through which a current is run in order to create a magnetic field. The magnetic field of the solenoid is the superposition of the fields due to the current through each coil. It is nearly uniform inside the solenoid and close to zero outside and is similar to the field of a bar magnet having a north pole at one end and a south pole at the other depending upon the direction of current flow
Experiment 3.9The Hall Effect and Magnetic Field of a Solenoid6Figure 3.9-2.B-field of a SolenoidThe field within the coil can be calculated from Ampere's Law:(3.9-18)B = μonlEquation (3.9-18) is really only valid for an infinitely long solenoid with closelyspaced turns. Near the ends of a finitelength solenoid, the field is smaller thanpredicted by (18); at the ends of a finite solenoid, B is 1/2 its maximum value. Thefactor of 1/2 can be understood from a symmetry argument: the end of a solenoid canbe thought of as an infinite solenoid with one half cut away. In an infinite solenoid,the total B-field at any point is due to the current in the coils to theleft and the currentin the coils to the right. By removing the coils on one side, the B-field is cut in halfExperimentalContents andProcedures1.HallEffect(1) Connect the wires of FB510A and FB510 according to color following the tab, andmakesurethattheHall probeelement iswellcentered intheHelmholtzcoils(2) Record a Gs/A (1 Gs=10-4 T) on the Helmholtz coils which can be used tocalculate the B-field in the centre: B = a - Im.(3)Recordb=4mm,d=0.5mm,L=3mm.(4) Measure Vh~ Is curve.@ Set IM= 100 mA;② Set Is = 0.50 mA, change the direction of B (IM) and Is (Ix), record the voltages ofVi,V2,V3and V,③ Set the current Is to the desired level 1.00 mA, 1.50 mA, 2.00 mA, ., 5.00 mArespectively, and record the voltages of Vi, V2, V3 and V4.(5) MeasureVH~IMcurve.① SetIs=1.00mA;② Set IM=100mA, change the direction ofB (Im)and Is (Ix),record thevoltages of
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 6 The field within the coil can be calculated from Ampere’s Law: 𝐵 = 𝜇0𝑛𝐼 (3.9-18) Equation (3.9-18) is really only valid for an infinitely long solenoid with closely spaced turns. Near the ends of a finitelength solenoid, the field is smaller than predicted by (18); at the ends of a finite solenoid, B is 1/2 its maximum value. The factor of 1/2 can be understood from a symmetry argument: the end of a solenoid can be thought of as an infinite solenoid with one half cut away. In an infinite solenoid, the total B-field at any point is due to the current in the coils to the left and the current in the coils to the right. By removing the coils on one side, the B-field is cut in half. Experimental Contents and Procedures 1. Hall Effect (1) Connect the wires of FB510A and FB510 according to color following the tab, and make sure that the Hall probe element is well centered in the Helmholtz coils. (2) Record a Gs/A (1 Gs=10-4 T) on the Helmholtz coils which can be used to calculate the B-field in the centre: 𝐵 = 𝑎 ∙ 𝐼𝑀. (3) Record b = 4 mm, d = 0.5 mm, L = 3 mm. (4) Measure VH ~ IS curve. ① Set IM = 100 mA; ② Set IS = 0.50 mA，change the direction of B (IM) and IS (Ix), record the voltages of 𝑉1, 𝑉2, 𝑉3 and 𝑉4; ③ Set the current IS to the desired level 1.00 mA, 1.50 mA, 2.00 mA, ., 5.00 mA respectively, and record the voltages of 𝑉1, 𝑉2, 𝑉3 and 𝑉4． (5) Measure VH ~ IM curve. ① Set IS = 1.00 mA; ② Set IM = 100 mA, change the direction of B (IM) and IS (Ix), record the voltages of B L I Figure 3.9-2. B-field of a Solenoid
Experiment 3.9The Hall Effect and Magnetic Field of a Solenoid7Vi, V2, V3 and V4;③ Set the current IM to the desired level 150 mA, 200 mA, ..., 500 mA respectivelyand record the voltages of Vi, V2, V3 and V4.(6) Measure conductivity ① Set IM=0mA;② Set Is= 0.20 mA, change the direction of Is (Ix), record the voltages of VAc+ andVAC-③ Set the current Is to the desired level 0.40 mA, 0.60 mA, 0.80 mA, 1.00 mArespectively,and record the voltages of VAc+ and VAc-.2. Magnetic Field of a Solenoid@ Connect the wires of FB510A and FB400 according to color following the tab,and make sure that the zero graduation line of Hall probe is aligned with the red line② Record KH (V/(A·T) ) at the right of the Hall probe. Set Is=4.00mA and IM= 400mA;③ Change the position of Hall probe from O to 17.o0 cm, and record the voltages ofVi,V2,V3andV4ateachpositionData Analysis and Results1.Hall Effect1).Calculate and list up Hall voltage under different conditions using Equation(3.9-17).2). Plot Vh versus IM, or VH versus Is. From the graph Hall voltage Vs. magnetic fieldcalculate Hall coefficient RH.3).Plot VAc versus Is, calculate conductivity 4).Calculate carrier concentration n and mobility μ.2. Magnetic Field of a Solenoid1). Calculate the Hall voltage at each position and give magnetic field B of thesolenoid.2). Plot B-x curve.3).Estimate the edge position of the solenoid
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 7 𝑉1, 𝑉2, 𝑉3 and 𝑉4; ③ Set the current IM to the desired level 150 mA, 200 mA, ., 500 mA respectively, and record the voltages of 𝑉1, 𝑉2, 𝑉3 and 𝑉4． (6) Measure conductivity σ ① Set IM = 0 mA; ② Set IS = 0.20 mA, change the direction of IS (Ix), record the voltages of VAC+ and VAC-. ③ Set the current IS to the desired level 0.40 mA, 0.60 mA, 0.80 mA, 1.00 mA respectively, and record the voltages of VAC+ and VAC-． 2. Magnetic Field of a Solenoid ① Connect the wires of FB510A and FB400 according to color following the tab, and make sure that the zero graduation line of Hall probe is aligned with the red line. ② Record KH (V/(A·T) ) at the right of the Hall probe. Set IS＝4.00mA and IM = 400 mA; ③ Change the position of Hall probe from 0 to 17.00 cm, and record the voltages of 𝑉1, 𝑉2, 𝑉3 and 𝑉4 at each position. Data Analysis and Results 1. Hall Effect 1). Calculate and list up Hall voltage under different conditions using Equation (3.9-17) ． 2). Plot VH versus IM, or VH versus IS. From the graph Hall voltage Vs. magnetic field calculate Hall coefficient RH． 3). Plot 𝑉𝐴𝐶 versus IS, calculate conductivity σ 4). Calculate carrier concentration n and mobility μ． 2. Magnetic Field of a Solenoid 1)．Calculate the Hall voltage at each position and give magnetic field B of the solenoid． 2)．Plot B-x curve． 3)．Estimate the edge position of the solenoid．
0Experiment 3.9 The Hall Effect and Magnetic Field of a SolenoidQuestions(1). Why the Hall voltage should be measured for both the directions of current aswell as ofmagnetic field?(2). How to measure the IR drop (Vo)?(3). How to calculate Vo using the voltages of Vi, V2, V3 and V? Please give theresults ofVounderdifferent Is
Experiment 3.9 The Hall Effect and Magnetic Field of a Solenoid 8 Questions (1). Why the Hall voltage should be measured for both the directions of current as well as of magnetic field? (2). How to measure the IR drop (V0)? (3). How to calculate 𝑉0 using the voltages of 𝑉1, 𝑉2, 𝑉3 and 𝑉4? Please give the results of 𝑉0 under different IS