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In SI units, permeability is measured in henries per meter H / m or H m − 1. Henry has the dimensions of [ M L 2 T − 2 A − 2 ] . Dimensions for magnetic permeability will be [ M L 2 T − 2 A − 2 ] / [ L ] = [ M L T − 2 A − 2 ]
The formula B 2 2 μ consists of the magnetic field intensity and the permeability of free space. The dimensional formula of the magnetic field is obtained as: F = q v B
The dimensional formula for permittivity of free space (ϵ 0) in the equation F = 1 4 π ϵ 0 q 1 q 2 r 2 where, symbols have their usual meaning is [M 1 L 3 A − 2 T − 4] [M − 1 L − 3 T 4 A 2] [M − 1 L − 3 A − 2 T − 4] [M 1 L 3 A 2 T − 4]
The dimensional formula of 1 2 μ 0 H 2 (μ 0 - Permeability of free space and H-magnetic field intensity) ...
Click here:point_up_2:to get an answer to your question :writing_hand:the dimensional formula for permeability mu is given by
Let [ϵ 0] denote the dimensional formula of the permittivity of vacuum and [μ 0] that of the permeability of vacuum. If M = mass, L = length, T = time, and I = electric current, then If M = mass, L = length, T = time, and I = electric current, then
Let [∈ 0] denote the dimensional formula of the permittivity of the vacuum and [μ 0] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current, then: Given F = 1 4 π ∈ 0 . q 1 q 2 r 2 , where q 1 and q 2 are charges and r is the distance.
Let [ε 0] denote the dimensional formula of the permittivity of the vaccum and [μ 0] denote the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current, then View Solution
The dimensions of ratio of magnetic flux (ϕ) and permeability (u) are: View Solution. Q 5.
Let [ε 0] denote the dimensional formula of the permittivity of the vaccum and [μ 0] denote the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current, then View Solution
