Question

An electron with charge e enters a uniform magnetic field b with a velocity v. The velocity is perpendicular to the magnetic field. The force on the charge e is given by. F= bev obtain the dimensions of b ans is L1M0t-2I-1

Answer 1

Answer:

B=[LMT^-2I^-1].

Explanation:

To find the dimension of the B or the magnetic field we have the dimension of force F= [MLT^-2], the dimension of the charge as [I] and the dimension of the velocity as [LT^-2].

So, [MLT^-2] = dimension of B *[I]*[LT^-2]. Since, formulae is F=BeV. So, on calculating we get that B= [MLT^-2]/[I]*[LT^-2]. Which on solving we will get B=[LMT^-2I^-1].

Answer 2

Answer:

see the attachment given above