Dimensions of permeability are

Options

A. `[A^(-2)M^(1)L^(1)T^(-2)]`
B. `[MLT^(-2)]`
C. `[ML^(0)T^(-1)]`
D. `[A^(-1)MLT^(-2)]`

Answer

To find the dimensions of permeability (μ), we can start from the definitions and relationships in electromagnetism. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the definition of permeability
Permeability (μ) is a measure of how easily a magnetic field can penetrate a material. It is often used in the context of solenoids and magnetic fields.

### Step 2: Use the formula for the magnetic field in a solenoid
The magnetic field (B) inside a solenoid is given by the formula:
\[ B = \mu \cdot n \cdot I \]
where:
- \( B \) is the magnetic field,
- \( \mu \) is the permeability,
- \( n \) is the number of turns per unit length (turns/meter),
- \( I \) is the current in amperes.

### Step 3: Rearranging the formula to find permeability
From the formula above, we can rearrange it to express permeability:
\[ \mu = \frac{B}{n \cdot I} \]

### Step 4: Identify the dimensions of each variable
- The dimension of magnetic field \( B \) is:
\[ [B] = \frac{F}{Q \cdot V} = \frac{M \cdot L \cdot T^{-2}}{A \cdot (L \cdot T^{-1})} = \frac{M}{A \cdot T} \]
- The dimension of \( n \) (number of turns per unit length) is:
\[ [n] = L^{-1} \]
- The dimension of current \( I \) is:
\[ [I] = A \]

### Step 5: Substitute the dimensions into the permeability equation
Now substitute the dimensions into the equation for permeability:
\[ [\mu] = \frac{[B]}{[n] \cdot [I]} = \frac{\frac{M}{A \cdot T}}{L^{-1} \cdot A} \]

### Step 6: Simplify the dimensions
This simplifies to:
\[ [\mu] = \frac{M}{A \cdot T} \cdot \frac{1}{L^{-1} \cdot A} = \frac{M \cdot L}{A^2 \cdot T} \]

### Final Result
Thus, the dimensions of permeability (μ) are:
\[ [\mu] = M \cdot L \cdot A^{-2} \cdot T^{-2} \]

### Conclusion
The correct option for the dimensions of permeability is:
\[ [\mu] = M^1 \cdot L^1 \cdot A^{-2} \cdot T^{-2} \]

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