Calculate speed v in crossfield hall effect

[th3plan]

Homework Statement

metal strip 6.66 cm long, 1.11 cm wide, and 0.837 mm thick moves with constant velocity through a uniform magnetic field B = 1.55 mT directed perpendicular to the strip, as shown in Fig. 28-37. A potential difference of 2.99 µV is measured between points x and y across the strip. Calculate the speed v.


http://img521.imageshack.us/img521/4905/wirept9.gif


Ok i know Fnet= qE+q(V x B) and then set equal to zero cause equilibrium and get E=-q x V

(the x means cross product)

So now explain to me Why E=Vb in this case. Then i use V=E/B to get my speed velocity. But i don't just understand why the negative sign is dropped of ? Is it because its absolute value or cause Electric Field is point from a + to a - potential, in the x-axis direction ?

Thanks for your help

 

[alphysicist]

Hi th3plan,

Homework Statement

metal strip 6.66 cm long, 1.11 cm wide, and 0.837 mm thick moves with constant velocity through a uniform magnetic field B = 1.55 mT directed perpendicular to the strip, as shown in Fig. 28-37. A potential difference of 2.99 µV is measured between points x and y across the strip. Calculate the speed v.


http://img521.imageshack.us/img521/4905/wirept9.gif


Ok i know Fnet= qE+q(V x B) and then set equal to zero cause equilibrium and get E=-q x V

(the x means cross product)

So now explain to me Why E=Vb in this case. Then i use V=E/B to get my speed velocity. But i don't just understand why the negative sign is dropped of ? Is it because its absolute value or cause Electric Field is point from a + to a - potential, in the x-axis direction ?

The sign is different because the equations are two different things. The equation with the minus sign is a vector equation (it should be \vec E = - \vec v\times\vec B); the other equation is only dealing with the magnitudes.

For example, suppose I am holding a weight W by applying a force F upwards. The vector equation for equilibrium would be

<br /> \vec F = -\vec W<br />
which means that my applied force is equal in magnitude and opposite in direction to the weight. If I wanted to calculate the magnitude of the force, I might write:

<br /> F = W<br />
which just means | \vec F | = | \vec W|.