A magnetising field of `1600 Am^(-1)` produces a magnetic flux `2.4 xx 10^(-5)` weber in an iron bar of cross-sectional area `0.2 cm^(2)` . The susceptibility of the iron bar will be

A. 1788
B. 1192
C. 596
D. 298

Options

A. 1788
B. 1192
C. 596
D. 298

Explanation

To find the susceptibility of the iron bar, we will follow these steps:

### Step 1: Note down the given values
- Magnetizing field (H) = 1600 Am⁻¹
- Magnetic flux (Φ) = 2.4 × 10⁻⁵ Wb
- Cross-sectional area (A) = 0.2 cm² = 0.2 × 10⁻⁴ m² = 2 × 10⁻⁶ m²

### Step 2: Calculate the magnetic field (B)
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