What is the magnitude and direction of the magnetic field?

Question

An electron experiences the greatest force as it travels 2.9x10^6 m/s in a magnetic field when it is moving northward . The force is upward and of magnitude 7.2x10^-13 N. What is the magnitude and direction of the magnetic field?

Answer 1

Using right hand rule (all physical quantities) B, F and V are vectors
Let unit vector along East be (+i), north (+j) and upwards (z-axis) be +k - also their negative counterparts in respective directions

given V = 2.9x10^6 ( j ) m/s northwards
F = 7.2x10^-13 ( k ) N along z-axis
|B| B-hat = ? where |B| is maagnitude of B and B-hat is resultant unit vector to be found

The force (F) on moving electron (V) in the magnetic field (B)is given by :

F = ( - e) [ V cross B] put vector values

7.2x10^-13 (k) = (-1.6x10^-19) [2.9x10^6 (j)]cross[ |B| B-hat ]

1.551 (k) = - [ j ] cross [ |B| B-hat ]

1.551 (k) = + [ |B| B-hat ] cross [ j ] >>> AXB = - BXA

k unit vector is equal to [ i cross j ]

1.551 [ i cross j ] = + [ |B| B-hat ] cross [ j ] ----(M)

This gives |B| = 1.551 Tesla (or Weber/m^2)

(M) clearly gives the direction of B along + i

B-hat = [ i ] that is field B will be directed towards EAST for the negatively charged ELECTRON. Had it been a positive charge obviously by right hand rule it would have been towards WEST.