JEE MainPhysicsUnits and Dimensions
Match List - I with List - II. List - I List - II (A) Magnetic permeability (I) [M L^2 T⁻² A⁻¹] (B) Magnetic flux (II) [M L T⁻² A⁻²] (C) Magnetic moment (III) [L⁻¹ A] (D) Magnetic intensity (IV) [L^2 A] Choose the correct answer from the options given below :
Options
- A(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
- B(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
- C(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
- D(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Correct answer
B. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Step-by-step solution
Let us find the dimensional formula for each quantity in List - I: (A) Magnetic permeability ( ): The force per unit length between two parallel currents is given by F L = ₀ I₁ I₂ 2 r . [ ] = [F][r] [L][I^2] = (M L T⁻²)(L) (L)(A^2) = [M L T⁻² A⁻²] . So, (A) matches with (II). (B) Magnetic flux ( ): From Faraday's law, induced emf e = - d dt . [ ] = [e][t] = [ Work / Charge ][t] = M L^2 T⁻² A T T = [M L^2 T⁻² A⁻¹] . So, (B) matches with (I). (C) Magnetic moment ( M ): Magnetic moment is current area. [M] = [I][A] =