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Match List - I with List – II {:("Li...

Match List - I with List – II
`{:("List - I","List - II"),("(A) Magnetic Induction", (ii) ML^(2)T^(-2)A^(-1)),("(B) Magnetic Flux", (ii) M^0L^(-1)A),("(C) Magnetic permeability",(iii)MT^(-2)A^(-1)),("(D) Magnetization ",(iv) MLT^(-2) A^(-2)):}`
Choose the most appropriate answer from the options given below

A

(A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)

B

(A)-(iii), (B)-(i), (C)-(iv), (D)-(ii)

C

(A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)

D

(A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of matching List - I with List - II, we will derive the dimensional formulas for each of the magnetic quantities in List - I and then match them with their corresponding formulas in List - II. ### Step-by-Step Solution: 1. **Magnetic Induction (B)**: - The formula for the magnetic induction \( B \) can be derived from the force on a current-carrying wire in a magnetic field: \[ F = I \cdot L \cdot B \] - Rearranging gives: \[ B = \frac{F}{I \cdot L} \] - The dimensions of force \( F \) are \( MLT^{-2} \), current \( I \) has dimensions \( A \), and length \( L \) has dimensions \( L \). - Therefore, substituting these into the equation gives: \[ B = \frac{MLT^{-2}}{A \cdot L} = \frac{MT^{-2}}{A} \] - Thus, the dimensional formula for magnetic induction \( B \) is: \[ [B] = MT^{-2}A^{-1} \] - This matches with option (iii) in List - II. 2. **Magnetic Flux (Φ)**: - The formula for magnetic flux \( Φ \) is given by: \[ Φ = B \cdot S \] - Where \( S \) is the area. The dimensions of area \( S \) are \( L^2 \). - Substituting the dimensions of \( B \): \[ Φ = (MT^{-2}A^{-1}) \cdot (L^2) = M L^2 T^{-2} A^{-1} \] - Thus, the dimensional formula for magnetic flux \( Φ \) is: \[ [Φ] = ML^2T^{-2}A^{-1} \] - This matches with option (i) in List - II. 3. **Magnetic Permeability (μ)**: - The magnetic permeability \( μ \) is defined as: \[ μ = \frac{B}{H} \] - We already have \( B \) and need to find \( H \) (magnetic field strength). The dimensions of \( H \) are: \[ H = \frac{B}{μ} \] - The dimensions of \( H \) are \( A L^{-1} \). - Substituting the dimensions of \( B \): \[ μ = \frac{MT^{-2}A^{-1}}{A L^{-1}} = \frac{MT^{-2}}{A^2} \] - Thus, the dimensional formula for magnetic permeability \( μ \) is: \[ [μ] = ML^{-1}T^{-2}A^{-2} \] - This matches with option (iv) in List - II. 4. **Magnetization (M)**: - The magnetization \( M \) is defined as: \[ M = \frac{magnetic \ dipole \ moment}{volume} \] - The dimensions of magnetic dipole moment are \( A L^2 \) and volume is \( L^3 \). - Thus, the dimensions of magnetization are: \[ M = \frac{A L^2}{L^3} = A L^{-1} \] - Thus, the dimensional formula for magnetization \( M \) is: \[ [M] = A L^{-1} \] - This matches with option (ii) in List - II. ### Final Matching: - (A) Magnetic Induction → (iii) \( MT^{-2}A^{-1} \) - (B) Magnetic Flux → (i) \( ML^2T^{-2}A^{-1} \) - (C) Magnetic Permeability → (iv) \( ML^{-1}T^{-2}A^{-2} \) - (D) Magnetization → (ii) \( A L^{-1} \) ### Answer: The correct matches are: - A → iii - B → i - C → iv - D → ii

To solve the problem of matching List - I with List - II, we will derive the dimensional formulas for each of the magnetic quantities in List - I and then match them with their corresponding formulas in List - II. ### Step-by-Step Solution: 1. **Magnetic Induction (B)**: - The formula for the magnetic induction \( B \) can be derived from the force on a current-carrying wire in a magnetic field: \[ F = I \cdot L \cdot B ...
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