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Question
An electron with charge e enters a uniform magnetic field `vec"B"` with a velocity `vec"v"`. The velocity is perpendicular to the magnetic field. The force on the charge is given by `|vec"F"|`= B e v.
Obtain the dimensions of `vec"B"`.
Solution
Given: `|vec"F"|`= B e v
Considering only magnitude, given equation is simplified to,
F = B e v
where, F = Force
B = Magnetic field
e = Charge
v = velocity
∴ B = `"F"/"e v"`
but, F = ma
∴ `["F"] = ["M"^1] xx ["L"^1"M"^0"T"^-2]`
∴ [F] = [L1M1T-2]
Electric charge, (e) = current × time
∴ [e] = [I1T1]
Velocity (v) = `"Displacement"/"time"`
∴ `["v"] = ["L"/"T"] = ["L"^1"M"^0"T"^-1]`
Now, [B] = `["F"/"e v"]`
`= (["L"^1"M"^1"T"^-2])/(["T"^1"I"^1]["L"^1"M"^0"T"^-1])`
∴ [B] = [L0M1T-2I-1]
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