Question

An electron with charge e enters a uniform.magnetic field B with a velocity v. Thevelocity is perpendicular to the magneticfield. The force on the charge e is givenbyF = Bev Obtain the dimensions of B.​

Answer 1

it is given that the magnetic force on the charge e , F = Bev

• dimension of magnetic force , F = [ML²T-²]
• dimension of charge , e = [AT]
• dimension of velocity, v = [LT-¹]

from dimensional analysis,

dimension of F = dimension of Bev

⇒ [ML-²T-²] = dimension of B × dimension of e × dimension of v

⇒ [ML²T-²] = dimension of B × [AT] × [LT-¹]

⇒ [ML²T-²] = dimension of B × [AL]

⇒ [ML²T-²]/[AL] = dimension of B

⇒ dimension of B = [MLT-²A-¹]

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