Trying to check my work on this homework problem Thanks A me

Trying to check my work on this homework problem. Thanks!

A metal strip lies on the x-y plane with current of 5.0 A flowing in the +y direction. The width of the strip (in the times direction) is 4.0 cm. A uniform B-field of 1.2 T is present along the +z direction. The charges move in the strip at the speed of 45 mu m/s. What is the hall voltage across the strip? Which side has a higher potential (L or R)? Why? The length of the strip is 12.0 cm. What is the force acting on the strip from the B-field and in which direction?

Solution

The current flowing in y direction and Magnetic Field is in z Direction. It means that electrons are moving in –y direction.

a) Force due to magnetic Field is given by

F₁= q (V_-y ×B_z)

Force due to Electric Field (Electric Field is due to induced Hall Voltage) is given by

F₂= qE

E= V_H/t

At some point inside the metal these forces will be balanced

F₁=F₂

This gives Hall Voltage

V_H= vBt

V_H= 216 mV

b) Since electron is moving in –y direction therefore cross product of (V_-y ×B_z) will give –x direction. It means that Side towards –x direction will be negative and that in +x direction will be positive.

c) Force on strip will be from –x direction only which is the result of cross product of (V_-y ×B_z).