Hall effect/Force on a current carrying wire

In summary, a conducting rectangular solid of dimensions dx=5m, dy=3m, dz=2m is moving at a constant velocity through a uniform magnetic field B=(30mT)j. The resulting electric field within the solid can be found by equating the electric force (qE) and magnetic force (Bqv) and solving for E. The potential difference across the solid can then be found using the equation E=V/d. However, the velocity of the solid is necessary to solve the problem accurately.
  • #1
Tanishq Nandan
122
5

Homework Statement


A conducting rectangular solid of dimensions
dx=5m
dy=3m
dz=2m
moves at a constant velocity through a uniform magnetic field B=(30mT)j. What are the resulting a)electric field within the solid,in unit vector notation?
b)potential difference across the solid?
20170709_222420-1.jpg

Homework Equations


●Electric Force=qE
●Magnetic Force=Bqv
●E=V/d
●Basic theory of Hall effect:On applying a magnetic field perpendicular to the block,some electrons get accumulated at one side of the block due to the magnetic force acting on them,which after a while causes an electric field to be produced inside the block across it's width ,which opposes and finally cancels the effect of the magnetic field.

The Attempt at a Solution


I think the velocity ought to be mentioned in the question.
It just says constant velocity.That could mean the block is moving either across the x-axis or along z axis. (Only then will Hall effect be observed and an electric field and hence potential difference will be observed)

Then,we can just equate electric and magnetic force (equations meantioned above)
And first find out E,then through the 3rd formula V.
However,I find out both E and V in terms of v,
to see if both the answers correspond to the same v as they should,and the answers matching(for a velocity of -200i,but we are not supposed to do that :p)
So,in a nutshell:Can I do this question without knowing the velocity before-hand??
 
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  • #2
Tanishq Nandan said:
So,in a nutshell:Can I do this question without knowing the velocity before-hand??
You are right, you need to know the velocity. I suspect that the velocity was meant to be given but was inadvertently left out.
 
  • #3
Ok,thanks
 

Related to Hall effect/Force on a current carrying wire

1. What is the Hall effect?

The Hall effect is a phenomenon where a magnetic field applied perpendicular to a current-carrying conductor produces a voltage difference across the conductor, perpendicular to both the current and the magnetic field. This effect was discovered by Edwin Hall in 1879.

2. How does the Hall effect work?

When a current-carrying wire is placed in a magnetic field, the free electrons in the wire experience a force due to their motion and the magnetic field. This creates a buildup of charges on one side of the wire, leading to a potential difference across the wire perpendicular to both the current and the magnetic field.

3. What are the applications of the Hall effect?

The Hall effect has a wide range of applications, including magnetic field measurement, current sensing, and the creation of Hall effect sensors for detecting the presence, strength, and direction of a magnetic field. It is also used in devices such as magnetometers, galvanometers, and Hall effect thrusters.

4. How is the Hall effect related to Lorentz force?

The Hall effect is a manifestation of the Lorentz force, which describes the force experienced by a charged particle moving in a magnetic field. In the case of the Hall effect, the Lorentz force acts on the free electrons in a current-carrying wire, causing them to move to one side of the wire and creating a potential difference.

5. What factors affect the Hall effect?

The Hall effect is affected by various factors, including the strength and direction of the magnetic field, the magnitude of the current, and the material properties of the conductor. The type of material, its thickness, and temperature can also influence the Hall effect. In addition, the geometry of the conductor and the angle between the current and magnetic field can affect the magnitude of the Hall voltage.

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