Question

Q.3 An electron with charge e enters a uniform magnetic field B → with a velocity v → . (2 Marks) The velocity is perpendicular to the magnetic field. The force on the charge e is given by, F =Bev → Obtain the dimensions of B → .

Answer 1

as it is given that,

F = BeV

where F is force , B is magnetic field, e is charge on electron and V is velocity.

so, dimension of B = dimension of {F/eV}

= dimension of F/{dimension of e × dimension of V}

we know,

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

then, dimension of B = [MLT-²]/[AT][LT-¹]

= [MLT-²]/[AL]

= [MT-²A-¹]

Answer 2

$\huge\bold\purple{Answer:-}$

= [MT-²A-¹]