Answer :

Given: The magnetic field density and the magnetic field strength

as related by the relation

where B is the magnetic field density, μm is the relative magnetic

permeability of the material and depends on the nature of the

material or the medium and H is the magnetic field strength which

tells upto what extent a material can be magnetized.

For free space, the relation becomes

…. (i)

where B0 is the field intensity in free space(vacuum), H0 is the

strength in free space and μ0 is the magnetic permeability of free

space. μ0 is a universal constant and its value is 4π × 10-7 T m A-1.

We can define a relation between the electric field and magnetic field

of an electromagnetic wave travelling in vacuum. It is

…. (ii)

where E0 is the electric field density and C is the speed of light in


From (i) and (ii)

…. (iii)

To find the dimension of , we will find the dimension of the

quantity in the R.H.S.

The dimension of C is [L T-1] because it is the speed.

To find the dimension of , we need to consider the Biot Savart’s

law which gives the magnetic field due to a current carrying

conductor. According to Biot Savart law

…. (iv)

where i is the current in the conductor, L is the length of conductor and R is the distance between conductor and point where field is to be found.

The scalar form of the above equation will be simply

…. (v)

From (v)

the dimensions of μ0 will be the dimensions of R.H.S

dimension of B= [M T-2 A �-1]

dimension of R2=[L2]

dimension of L=[L]

dimension of i=[A]

dimension of R.H.S

Therefore dimension of μ0 is .

Now the dimension of R.H.S of eq(iii) will be dimension of C × dimension of μ0

dimension of

As the dimensions of electric resistance given by Ohm’s Law(V=iR) is

Therefore has the same dimensions as that of electrical resistance.

Also the value of is a constant because the R.H.S of eq(iii) is the

product of 2 universal constants.

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