Force on a charge on magnetic field

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Homework Help Overview

The discussion revolves around a problem in magnetism, specifically concerning the force experienced by a charged particle (an electron) moving in a magnetic field. The original poster is trying to determine both the magnitude and direction of this force given specific parameters, including the velocity of the electron and the strength of the magnetic field.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the right-hand rule to determine the direction of the force but expresses confusion regarding the correct application due to the negative charge of the electron. Other participants suggest visualizing the force direction through vector rotation and checking the calculations mathematically.

Discussion Status

Participants are actively engaging in clarifying the application of the right-hand rule and exploring the mathematical aspects of the vector product involved in calculating the force. Some guidance has been offered regarding the use of the determinant method for vector products, although the original poster has not yet confirmed understanding of this method.

Contextual Notes

The original poster has indicated a misunderstanding in the application of the right-hand rule and is seeking further clarification on the mathematical approach to verify their answer. There is an acknowledgment of potential errors in orientation and understanding of the concepts involved.

Genericcoder
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Hi guys,

I am solving some magnetism questions,but somewhat confused about right hand rule.

Determine the magnitude and direction of the force on an electron traveling 8.75 * 10^5 m/s horizontally to the east in vertically upward magnetic field of strnegth 0.75T?


We know since this is perpindcular to the magnetic field it will have maximum since sin(90) would be 1,so we can calculate the the force easily.

F = qVBsin;
F = 1.05;

What I don't understand is the direction first I did normally right hand rule,but since that charge is an electron it should be the opposite to what right hand rule gives us.

So I got the direction as South,but my book says west that should be the direction on a given proton not electron right ?
 
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You can visualize the direction of the force if you imagine that you rotate v into B (by the angle less than pi). The force points to that direction from where the rotation looks anti-clockwise. To the South for a proton, opposite for the electron, too North. You did something wrong with the right-hand rule. F=q vxB (vector product).

ehild
 

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I see I just rechecked I orientated by hand wrong that's why I got wrong answer :S.
Is their a way to check if I got my answer correctly mathematically ?
 
You can find the force from the components of v and B. Do you know the determinant method for calculating a vector product?

ehild
 
No I don't know about it :S could you suggest a tutorial to read about this method?
 
The determinant only makes easier to memorize cross product. (See for example http://mathworld.wolfram.com/CrossProduct.html).

You certainly know that

[itex]\vec a\times \vec b=(a_yb_z-a_zb_y) \vec i+(a_zb_x-a_xb_z) \vec j+(a_xb_y-a_yb_x) \vec k[/itex].

The force is [itex]\vec F= q\left[\vec v \times \vec B\right][/itex]. If the x-axis points to East, the y-axis points to North and the z axis point upward, [itex]\vec v=v \vec i[/itex] and [itex]\vec B=B \vec k[/itex], and [itex]\vec F=q v \vec i\times B \vec k=-qvB \vec j[/itex].

ehild
 
I see thanks a lot :).
 

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