AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

The permeability of vacuum can be expressed as,
µ0∝eambcchd
‌‌=keambcchd
Here, k is the proportionality constant.
Substituting the dimensions of different physical quantities in the above expression, we get
[MLT−2A−2]=[AT]a[M]b[LT−1]c[ML2T−1]d
[MLT−2A−2]=[Mb+dLc+2dTa−c−dAa]
The values of a,b c and d is calculated by comparing the power on both sides.
a‌‌=−2
b+d‌‌=1
c+2d‌‌=1
a−c−d‌‌=−2
On solving the above equations we get,
a=−2
b=0
c=−1
d=1
So, the permeability of free space can be expressed as,
µ0=‌

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