Question

An electron with charge e enters q uniform magnetic field B vector with a velocity v vector. The velocity is perpendicular to the magnetic field. The force on the charge e is given by magnitude off vector =Bev obtain the dimension of B vector

Answer 1

Answer:

$[M^1T^{-2}A^{-1}]$

Explanation:

We are given that force on the charge e is given b y the magnitude of vector

F=Bev

We have to find the dimensions of B vector

We know that dimension of Force=$[M^1LT^{-2}]$

Dimension of charge=$[AT]$

Dimension of velocity=$[LT^{-1}]$

Substitute  all values then we get

$B=\frac{F}{ev}$

$B=\frac{[MLT^{-2}]}{[AT]\cdot[LT^{-1}]}$

$B=[M^1T^{-2}A^{-1}]$

Hence, the dimension of B=$[M^1T^{-2}A^{-1}]$