JEE Physics 2026 Question with Solution

Given: Match magnetic quantities with their dimensional formulae.

Find: The correct correspondence between LIST-I and LIST-II.

Step 1: Magnetic induction $B$. Magnetic induction is force per unit current per unit length.

$$[B] = M L^{0} T^{-2} A^{-1}$$

Hence, A $\rightarrow$ III.

Step 2: Magnetic flux $\Phi$. Magnetic flux is given by $B \times \text{area}$.

$$[\Phi] = (M L^{0} T^{-2} A^{-1}) \times L^{2} = M L^{2} T^{-2} A^{-1}$$

Hence, B $\rightarrow$ IV.

Step 3: Magnetic permeability $\mu$. From $B = \mu H$,

$$[\mu] = \frac{[B]}{[H]} = \frac{M L^{0} T^{-2} A^{-1}}{A L^{-1}} = M L T^{-2} A^{-2}$$

Hence, C $\rightarrow$ I.

Step 4: Self inductance $L$. Self inductance is flux per unit current.

$$[L] = \frac{M L^{2} T^{-2} A^{-1}}{A} = M L^{2} T^{-2} A^{-2}$$

Hence, D $\rightarrow$ II.

Conclusion: The correct matching is A-III, B-IV, C-I, D-II. Therefore, the correct option is D.

Given: The quantities are magnetic induction, magnetic flux, magnetic permeability, and self inductance.

Find: Their dimensional matches.

Use basic definitions instead of memorizing formulae.

For magnetic induction,

$$B = \frac{F}{I \, l}$$

So,

$$[B] = \frac{M L T^{-2}}{A \, L} = M L^{0} T^{-2} A^{-1}$$

Therefore, A matches III.

For magnetic flux,

$$\Phi = B \times \text{area}$$

Thus,

$$[\Phi] = (M L^{0} T^{-2} A^{-1}) \times L^{2} = M L^{2} T^{-2} A^{-1}$$

Therefore, B matches IV.

For magnetic permeability, use

$$B = \mu H$$

and magnetic field strength has dimensions

$$[H] = A L^{-1}$$

Hence,

$$[\mu] = \frac{M L^{0} T^{-2} A^{-1}}{A L^{-1}} = M L T^{-2} A^{-2}$$

Therefore, C matches I.

For self inductance,

$$L = \frac{\Phi}{I}$$

So,

$$[L] = \frac{M L^{2} T^{-2} A^{-1}}{A} = M L^{2} T^{-2} A^{-2}$$

Therefore, D matches II.

Hence the final arrangement is A-III, B-IV, C-I, D-II, which corresponds to option D.

Note: The raw option numbering on the source corresponds to option (4), and that is labeled D here.