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JEE MAIN 2026 · Units and Measurements

JEE Main 2026 Session 2 Question Paper: 8 Apr, Shift 2 (B.E./B.Tech), question 27

Updated
Single correct+4 / -1Answer matches NTA’s final key

Consider the equation H=xpϵqErtsH = \dfrac{x^p \epsilon^q E^r}{t^s}

where HH = magnetic field; EE = electric field, ϵ\epsilon = permittivity, xx = distance, tt = time.

The values of pp, qq, rr and ss respectively are:

Choose one

Solution

Solution

Here HH is the magnetising field, measured in A m−1^{-1}, so [H]=[A L−1][H] = [\mathrm{A\,L^{-1}}].

The quickest route uses a known combination: ϵE\epsilon E is the electric displacement, in C m−2^{-2} =[A T L−2]= [\mathrm{A\,T\,L^{-2}}]. Then

x ϵEt=[L][A T L−2][T]=[A L−1]=[H].\frac{x\,\epsilon E}{t} = \frac{[\mathrm L][\mathrm{A\,T\,L^{-2}}]}{[\mathrm T]} = [\mathrm{A\,L^{-1}}] = [H].

(This is the displacement-current form of Ampère's law: H∼x⋅d(ϵE)dtH \sim x \cdot \frac{d(\epsilon E)}{dt}.) The dimensions balance with p=q=r=s=1p = q = r = s = 1.

Answer: 1, 1, 1, 1.

Answer checked against the official answer key. The solution is our own working.

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