A conductor `ab` of arbitrary shape carries current I flowing from `b` to `a`. The length vector `ab` is oriented from `a` to `b`. The force `F` experienced by this conductor in as uniform magnetic field B is

Options

A. `F=-I(abxxB)`
B. `F=I(bxxab)`
C. `F=I(baxxB)`
D. All of the above

Explanation

To solve the problem of finding the force experienced by a conductor `ab` carrying current `I` in a uniform magnetic field `B`, we can follow these steps:

### Step 1: Understand the Direction of Current and Length Vector
The current `I` is flowing from point `B` to point `A`, while the length vector `\(\vec{L}\)` is oriented from `A` to `B`. Therefore, we can express the length vector as:
\[ \vec{L} = \vec{AB} = \vec{b} - \vec{a} \]

### Step 2: Use the Formula for Magnetic Force
The force `\(\vec{F}\)` experienced by a current-carrying conductor in a magnetic field is given by the formula:
\[ \vec{F} = I \vec{L} \times \vec{B} \]
Where:
- `I` is the current,
- `\(\vec{L}\)` is the length vector of the conductor,
- `\(\vec{B}\)` is the magnetic field.

### Step 3: Substitute the Length Vector
Since the current is flowing from `B` to `A`, we can express the length vector as:
\[ \vec{L} = \vec{AB} = -(\vec{BA}) \]
Thus, we can rewrite the force as:
\[ \vec{F} = I (-\vec{BA}) \times \vec{B} \]
This simplifies to:
\[ \vec{F} = -I \vec{BA} \times \vec{B} \]

### Step 4: Interpret the Cross Product
The cross product `\(\vec{BA} \times \vec{B}\)` gives a vector that is perpendicular to both `\(\vec{BA}\)` and `\(\vec{B}\)`. The negative sign indicates that the direction of the force is opposite to that of the vector `\(\vec{BA} \times \vec{B}\)`.

### Step 5: Conclusion
The final expression for the force experienced by the conductor `ab` is:
\[ \vec{F} = -I \vec{BA} \times \vec{B} \]
This indicates that the force is dependent on the current `I`, the length of the conductor, and the magnetic field `B`.

### Summary of the Solution
The force experienced by the conductor `ab` in a uniform magnetic field `B` is given by:
\[ \vec{F} = I \vec{AB} \times \vec{B} = -I \vec{BA} \times \vec{B} \]

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