Units & Dimensions

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18 of 149

Question 18 of 149

Consider the equation H=xpϵqErtsH = \dfrac{x^p \epsilon^q E^r}{t^s}, where H=H= magnetic field; E=E= electric field, ϵ=\epsilon= permittivity, x=x= distance, t=t= time. The values of p,q,rp, q, r and ss respectively are:

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Answer: (A) 1,1,1,11, 1, 1, 1

The dimensional formulas for the given physical quantities are:

Magnetic field intensity [H]=[M0L−1T0A1][H] = [M^0 L^{-1} T^0 A^1]

Distance [x]=[M0L1T0A0][x] = [M^0 L^1 T^0 A^0]

Permittivity [ϵ]=[M−1L−3T4A2][\epsilon] = [M^{-1} L^{-3} T^4 A^2]

Electric field [E]=[M1L1T−3A−1][E] = [M^1 L^1 T^{-3} A^{-1}]

Time [t]=[M0L0T1A0][t] = [M^0 L^0 T^1 A^0]

Given the equation H=xpϵqErtsH = \dfrac{x^p \epsilon^q E^r}{t^s}, we can write it in terms of dimensions as:

[M0L−1T0A1]=[L]p[M−1L−3T4A2]q[MLT−3A−1]r[T]−s[M^0 L^{-1} T^0 A^1] = [L]^p [M^{-1} L^{-3} T^4 A^2]^q [M L T^{-3} A^{-1}]^r [T]^{-s}

Combining the powers of M,L,TM, L, T, and AA on the right side:

[M0L−1T0A1]=[M−q+rLp−3q+rT4q−3r−sA2q−r][M^0 L^{-1} T^0 A^1] = [M^{-q+r} L^{p-3q+r} T^{4q-3r-s} A^{2q-r}]

Equating the powers of corresponding fundamental quantities from both sides:

For MM: 0=−q+r  ⟹  q=r0 = -q + r \implies q = r

For AA: 1=2q−r1 = 2q - r

Substituting r=qr = q into the equation for AA:

1=2q−q  ⟹  q=11 = 2q - q \implies q = 1

Since q=rq = r, we get r=1r = 1.

For LL: −1=p−3q+r-1 = p - 3q + r

Substituting q=1q = 1 and r=1r = 1:

−1=p−3(1)+1  ⟹  −1=p−2  ⟹  p=1-1 = p - 3(1) + 1 \implies -1 = p - 2 \implies p = 1

For TT: 0=4q−3r−s0 = 4q - 3r - s

Substituting q=1q = 1 and r=1r = 1:

0=4(1)−3(1)−s  ⟹  0=1−s  ⟹  s=10 = 4(1) - 3(1) - s \implies 0 = 1 - s \implies s = 1

Thus, the values are p=1p = 1, q=1q = 1, r=1r = 1, and s=1s = 1.

Answer: 1,1,1,11, 1, 1, 1

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