Question

5.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] = B e v. Obtain the dimensions of B.​

Answer 1

Explanation:

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: [L'MºT21-11.

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Answer 2

$[MT^{-2}A^{-1}]$

Explanation:

Given:

• A charge 'e'

• A uniform magnetic field 'B'

• Velocity of charge 'v'

• Velocity is perpendicular to the magnetic field

To find:

Dimensional formula of 'B'

Solution:

As we know, force on the charge 'e' entering a uniform magnetic field 'B' with a velocity 'v' is given by the formula :

$\bf \underline {\boxed {F=Bev}}$

$\implies \bf B\:=\:\frac{F}{ev}...(1)$

Now,

• F = ma

Dimensional formula of 'F' = [MLT⁻²]

• e = I t

Dimensional formula of 'e' = [AT]

• v = s/t

Dimensional formula of 'v' = [LT⁻¹]

Substituting above dimensional formulas in eq.(1)

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

$\implies B=[ML^0A^{-1}T^{-2}]$

$\implies \bf \boxed{B=[MT^{-2}A^{-1}]}$