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Ampere's Force

CSCA Ampere's Force study guide organized around the publicly available CSCA syllabus. Practice Physics questions on aicsca.com.

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Topic: Electromagnetism > Magnetic Field > Ampere's Force

1. Core Concept: Ampere's Force

**Ampere's Force** is the force exerted by a magnetic field on a **current-carrying wire**. It is the macroscopic manifestation of the Lorentz force and the driving principle behind electric motors.

**CSCA Exam Focus**:

* Applicable conditions for $F=BIL$.

* Using the **Left-Hand Rule** (3D spatial visualization).

* Solving **Equilibrium** and **Dynamics** problems involving Ampere's force.

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2. Core Formulas

(1) Magnitude

$$F = B I L \sin\theta$$

* **$B$**: Magnetic Flux Density (T).

* **$I$**: Current (A).

* **$L$**: **Effective Length** of the wire (m).

* *Straight wire*: Actual length inside the field.

* *Curved wire*: The **straight-line distance** between the two endpoints (in a uniform field).

* **$\theta$**: Angle between **Current direction** and **B-field direction**.

**Special Cases**:

* **$I \perp B$** ($\theta = 90^\circ$): $F = BIL$ (**Max**).

* **$I \parallel B$** ($\theta = 0^\circ$): $F = 0$ (**Min**).

(2) Direction: Left-Hand Rule

Distinguish carefully: **Left Hand for Force**, Right Hand for Field generation.

* **Step 1**: Open your **Left Hand**, thumb $\perp$ fingers.

* **Step 2**: Let Magnetic Field lines ($B$) **pierce the palm**.

* **Step 3**: Fingers point in the direction of **Current** ($I$).

* **Step 4**: **Thumb** points to **Ampere's Force** ($F$).

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3. Problem-Solving Models

Model 1: 2D Cross-Section Analysis

Convert 3D diagrams into **2D cross-sections**.

* Current: $\odot$ (Out of page), $\otimes$ (Into page).

* Field: $\times$ (In), $\cdot$ (Out), or Arrows.

Model 2: Effective Length Method

For curved wires (e.g., a semi-circle), use the **chord length** connecting the ends as $L$.

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4. Typical Examples

**Example 1: Basic Calculation**

**Problem**: Wire $L=0.5m$, $I=4A$, in field $B=0.2T$.

(1) If $I \perp B$, find $F$.

(2) If angle between $I$ and $B$ is $30^\circ$, find $F$.

**Solution**:

(1) $F = BIL = 0.4 N$.

(2) $F = BIL \sin30^\circ = 0.2 N$.

**Example 2: Equilibrium on Inclined Plane (Tricky)**

**Problem**: Smooth slope $\theta=30^\circ$. Wire mass $m=0.01kg$, $L=0.2m$, $I=5A$ (Into page). Field $B=0.4T$ is **Vertically Upward**. Determine motion ($g=10m/s^2$).

**Solution**:

1. **Direction of F**:

* $B$ Up (Palm down), $I$ In (Fingers in) $\rightarrow$ **Thumb points Right**. Force is **Horizontal Right**.

2. **Magnitude**: $F = BIL = 0.4 N$ (since $B \perp I$).

3. **Decomposition (Along slope)**:

* Gravity component (down slope): $mg \sin30^\circ = 0.05 N$.

* Ampere component (up slope): Force is horizontal, angle with slope is $30^\circ$. $F_x = F \cos30^\circ \approx 0.346 N$.

4. **Result**:

Upward force ($0.346N$) > Downward force ($0.05N$).

The wire accelerates **up the slope**.

5. **Acceleration**: $a = (0.346 - 0.05) / 0.01 = 29.6 m/s^2$.

**Example 3: Parallel Wires**

**Problem**: Two parallel wires carry current in the **same** direction. Do they attract or repel?

**Solution**: **Attract**. (Same direction attracts, Opposite repels).

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5. CSCA Common Pitfalls

1. **Meaning of $L$**: Always find the "Effective Length". A closed loop in a uniform field has net force = 0.

2. **Drawing Diagrams**: Always draw a **Side View (Cross-section)** for inclined plane or rail problems.

3. **Angles**: When decomposing horizontal forces onto a slope, ensure you use the correct trig function (usually $\cos\theta$ for horizontal force along the slope).