`("Weber" xx "ampere")/("meter")` is equal to .

Options

A. joule
B. newton
C. henry
D. watt

Explanation

To solve the problem of determining what \((\text{Weber} \times \text{Ampere})/\text{meter}\) is equal to, we will break it down step by step.

### Step 1: Understand the Units
We start by identifying the units involved in the expression.
- Weber (Wb) is the unit of magnetic flux.
- Ampere (A) is the unit of electric current.
- Meter (m) is the unit of length.

### Step 2: Write the Units in Fundamental Form
Weber can be expressed in terms of fundamental units:
1 Weber = \(1 \, \text{kg} \cdot \text{m}^2/\text{s}^2/\text{A}\)

So, we can rewrite the expression:
\[ \text{Weber} \times \text{Ampere} = \text{Wb} \times \text{A} = \left( \frac{\text{kg} \cdot \text{m}^2}{\text{s}^2 \cdot \text{A}} \right) \times \text{A} \]

### Step 3: Simplify the Expression
Now, we can simplify the expression:
\[ \text{Wb} \times \text{A} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^2} \]

Next, we divide by meter:
\[ \frac{\text{Wb} \times \text{A}}{\text{m}} = \frac{\frac{\text{kg} \cdot \text{m}^2}{\text{s}^2}}{\text{m}} = \frac{\text{kg} \cdot \text{m}^2}{\text{s}^2 \cdot \text{m}} = \frac{\text{kg} \cdot \text{m}}{\text{s}^2} \]

### Step 4: Identify the Resulting Unit
The resulting unit \(\frac{\text{kg} \cdot \text{m}}{\text{s}^2}\) is recognized as the unit of force, which is Newton (N).

### Conclusion
Thus, the expression \((\text{Weber} \times \text{Ampere})/\text{meter}\) is equal to Newton.

### Final Answer
The answer is **Newton**.

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