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The dimensions [ M L T − 2 A − 2 ] [MLT^{-2}A^{-2}] [ M L T − 2 A − 2 ] belong to:
medium
Magnetism and Matter
2022
physics
Explanation To solve this problem, we need to determine which physical quantity has the dimensions [ M L T − 2 A − 2 ] . [MLT^{-2}A^{-2}]. \\
[ M L T − 2 A − 2 ] . Let's analyze the dimensions of each option: \\
1. Magnetic flux ( Φ ) : (\Phi): \\
( Φ ) :
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The formula for magnetic flux is
Φ = B ⋅ A , \Phi = B \cdot A, Φ = B ⋅ A , where
is magnetic field and
is area.
The dimensions of magnetic field
are
[ M T − 2 A − 1 ] , [MT^{-2}A^{-1}], [ M T − 2 A − 1 ] , and area
has dimensions
Thus, the dimensions of magnetic flux are
[ M T − 2 A − 1 ] ⋅ [ L 2 ] = [ M L 2 T − 2 A − 1 ] . [MT^{-2}A^{-1}] \cdot [L^2] = [ML^2T^{-2}A^{-1}]. \\
[ M T − 2 A − 1 ] ⋅ [ L 2 ] = [ M L 2 T − 2 A − 1 ] . 2. Self inductance
The formula for self inductance is
L = Φ I , L = \frac{\Phi}{I}, L = I Φ , where
is magnetic flux and
is current.
Using the dimensions of magnetic flux from above,
[ M L 2 T − 2 A − 1 ] , [ML^2T^{-2}A^{-1}], [ M L 2 T − 2 A − 1 ] , and current
with dimensions
the dimensions of self inductance are
[ M L 2 T − 2 A − 1 ] [ A ] = [ M L 2 T − 2 A − 2 ] . \frac{[ML^2T^{-2}A^{-1}]}{[A]} = [ML^2T^{-2}A^{-2}]. \\
[ A ] [ M L 2 T − 2 A − 1 ] = [ M L 2 T − 2 A − 2 ] . 3. Magnetic permeability
The formula for magnetic permeability is
μ = B H , \mu = \frac{B}{H}, μ = H B , where
is magnetic field and
is magnetic field strength.
The dimensions of magnetic field strength
are
[ L − 1 A ] . [L^{-1}A]. \\
[ L − 1 A ] . Thus, the dimensions of magnetic permeability are
[ M T − 2 A − 1 ] [ L − 1 A ] = [ M L T − 2 A − 2 ] . \frac{[MT^{-2}A^{-1}]}{[L^{-1}A]} = [MLT^{-2}A^{-2}]. \\
[ L − 1 A ] [ M T − 2 A − 1 ] = [ M L T − 2 A − 2 ] . 4. Electric permittivity
( ε ) : (\varepsilon): \\
( ε ) : The formula for electric permittivity is
ε = D E , \varepsilon = \frac{D}{E}, ε = E D , where
is electric displacement and
is electric field.
The dimensions of electric displacement
are
[ L − 2 A T ] , [L^{-2}AT], [ L − 2 A T ] , and electric field
has dimensions
[ M L T − 3 A − 1 ] . [MLT^{-3}A^{-1}]. \\
[ M L T − 3 A − 1 ] . Thus, the dimensions of electric permittivity are
[ L − 2 A T ] [ M L T − 3 A − 1 ] = [ M − 1 L − 3 T 4 A 2 ] . \frac{[L^{-2}AT]}{[MLT^{-3}A^{-1}]} = [M^{-1}L^{-3}T^4A^2]. \\
[ M L T − 3 A − 1 ] [ L − 2 A T ] = [ M − 1 L − 3 T 4 A 2 ] . From the above analysis, the dimensions
[ M L T − 2 A − 2 ] [MLT^{-2}A^{-2}] [ M L T − 2 A − 2 ] correspond to magnetic permeability.
Therefore, the correct option is 3: Magnetic permeability.