Question

to find dimension of radius of magnetic field

Answer 1

Dimensions of L, MT−2L2A−2

Dimensions of R,

ML2T−3A−2

Explanation:

Firstly consider resistance.

It's defining equation is, Ohm's law,

V=IR
⇒R=VI

Now V has units of (electric field)*(distance).

But electric field has units (force)/(charge).

Also, charge has dimensions of (current)(time) and force has dimensions (mass)(length)/(time)^2.

Thus, dimensions of V is,

[V]=LMLT−2AT
⇒[V]=ML2T−3A−1

Current I has dimensions [I]=A

Thus, dimensions of resistance,

[R]=[V][I]=ML2T−3A−2

For inductance, the defining equation is,

ϕ=LI

But ϕ has units (magnetic field)*(length)^2

Magnetic field from Lorentz force law has units, (Force)(velocity)^(-1)(charge)^(-1)

Therefore, dimensions of magnetic field,

[B]=MLT−2LT−1AT
⇒[B]=MLT−2LA
⇒[B]=MT−2A−1

Therefore dimensions of magnetic flux,

[ϕ]=[B]L2
⇒[ϕ]=MT−2L2A−1

Thus finally, dimensions of inductance,

[L]=[ϕ][I]
⇒[L]=MT−2L2A−2

hope it will help you. I don't know exact answer but I know that it will help you.

Answer 2

Answer:

We have the equation,

F=qvBsinθ

Dimension of Force, F=[MLT−2 ]

Dimension of Velocity, v=[LT−1 ]

Dimension of Charge, q=[AT]

Sinθ is dimensionless

Then,

Magnetic Field, B= Force/velocity × charge

= [MLT−2 ]/ [LT-1][AT]

=[MT−2A−1 ]