Are you struggling to grasp the complex world of electromagnetism? Understanding Magnetic Fields (चुंबकीय क्षेत्र) is essential for students in Class 12, engineering, and physics enthusiasts. This comprehensive guide breaks down the core concepts and the two fundamental laws—the Biot-Savart Law and Ampère’s Circuital Law—that govern the generation of these fields. Master these topics to score higher in exams and build a solid physics foundation!
1. Understanding Magnetic Fields (चुंबकीय क्षेत्र को समझना)
A Magnetic Field is a region around a magnetic material or a moving electric charge where a force would be felt by another magnetic material or a moving charge.
- Generation (उत्पत्ति): A magnetic field is primarily generated by moving electric charges (electric current). This is the key link between electricity and magnetism.
- Vector Quantity (सदिश राशि): It is denoted by the vector $\mathbf{B}$ and measured in Tesla (T) or Gauss (G) ($1 \text{ T} = 10^4 \text{ G}$).
Hindi Translation: चुंबकीय क्षेत्र किसी चुंबक या गतिमान विद्युत आवेश के चारों ओर का वह क्षेत्र है जिसमें किसी अन्य चुंबक या गतिमान आवेश द्वारा बल महसूस किया जाता है। इसका मुख्य स्रोत गतिमान विद्युत आवेश (विद्युत धारा) है। इसे $\mathbf{B}$ से दर्शाया जाता है और यह एक सदिश राशि है।
2. The Biot-Savart Law: The Fundamental Tool (बायो-सावर्ट का नियम: मूलभूत उपकरण)
The Biot-Savart Law is a mathematical tool used to determine the magnetic field ($\mathbf{B}$) at any point in space due to a small segment of current-carrying wire, known as a current element ($I d\mathbf{l}$).
💡 The Mathematical Statement
The magnitude of the magnetic field $d\mathbf{B}$ produced by a current element $I d\mathbf{l}$ at a distance $r$ is:
$$|d\mathbf{B}| = \frac{\mu_0}{4\pi} \frac{I d l \sin\theta}{r^2}$$
Where:
- $\mu_0$ is the permeability of free space (निर्वात की चुंबकशीलता).
- $I dl$ is the current element (धारा अवयव).
- $r$ is the distance from the element to the point of observation.
- $\theta$ is the angle between the current element $d\mathbf{l}$ and the position vector $\mathbf{r}$.
📌 Key Features
- Analogy to Coulomb’s Law: It’s similar to Coulomb’s Law for electric fields, as it is an inverse-square law ($\propto 1/r^2$).
- Superposition Principle: The total magnetic field is found by integrating the contribution of all current elements: $\mathbf{B} = \int d\mathbf{B}$.
- Direction: The direction of $\mathbf{B}$ is perpendicular to both $d\mathbf{l}$ and $\mathbf{r}$ (found using the Right-Hand Thumb Rule).
Hindi Translation: बायो-सावर्ट का नियम किसी धारा अवयव ($I d\mathbf{l}$) के कारण किसी बिंदु पर उत्पन्न चुंबकीय क्षेत्र ($d\mathbf{B}$) को निर्धारित करता है। यह कूलम्ब के नियम के समान ही व्युत्क्रम वर्ग नियम का पालन करता है। कुल चुंबकीय क्षेत्र ज्ञात करने के लिए सभी अवयवों के योगदान का समाकलन (Integration) किया जाता है।
3. Ampère’s Circuital Law: The Shortcut (एम्पीयर का परिपथीय नियम: त्वरित विधि)
Ampère’s Circuital Law is an alternative and often simpler method to calculate magnetic fields, particularly when the current distribution has high symmetry (like a long straight wire or a solenoid).
📐 The Statement
The line integral of the magnetic field ($\mathbf{B}$) around any closed loop (called the Ampèrian loop) is equal to $\mu_0$ times the total current ($I_{\text{enclosed}}$) threading through the loop.
$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}$$
Where:
- $\oint \mathbf{B} \cdot d\mathbf{l}$ is the line integral of $\mathbf{B}$ over the closed loop.
- $I_{\text{enclosed}}$ is the net current passing through the area enclosed by the loop.
⚡ Why is Ampère’s Law Useful?
- Symmetry (समरूपता): It drastically simplifies calculations for highly symmetric cases (e.g., finding $\mathbf{B}$ due to an infinite straight wire is much easier with Ampère’s Law than with the Biot-Savart Law).
- Efficiency: It serves as the magnetic analogue to Gauss’s Law in electrostatics.
Hindi Translation: एम्पीयर का परिपथीय नियम उच्च समरूपता वाले मामलों में चुंबकीय क्षेत्र की गणना के लिए एक सरल तरीका प्रदान करता है। इसके अनुसार, किसी बंद लूप (एम्पीरियन लूप) के चारों ओर चुंबकीय क्षेत्र ($\mathbf{B}$) का रेखा समाकलन लूप से घिरी हुई कुल धारा ($I_{\text{enclosed}}$) के $\mu_0$ गुना के बराबर होता है।
4. Biot-Savart Law vs. Ampère’s Law (तुलना)
| Feature (विशेषता) | Biot-Savart Law (बायो-सावर्ट नियम) | Ampère’s Circuital Law (एम्पीयर नियम) |
| Applicability (लागू) | Universal (सार्वभौमिक), for any current element. | Limited to cases with high symmetry. |
| Calculation (गणना) | Requires complex integration over the entire current path. | Reduces integration to a simple algebraic equation for symmetric paths. |
| Analogy (तुलना) | Analogous to Coulomb’s Law in electrostatics. | Analogous to Gauss’s Law in electrostatics. |
📈 Ready to Ace Your Electromagnetism Section?
Understanding the differences and applications of these two laws is the key to solving complex problems related to magnetic fields. Practice using the Biot-Savart Law for complex geometries and Ampère’s Law for symmetric ones!
Applications of Ampère’s Circuital Law: Solenoid and Toroid (एम्पीयर के नियम के अनुप्रयोग)
Ampère’s Circuital Law ($\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}$) is incredibly powerful for determining the magnetic field ($\mathbf{B}$) produced by current-carrying conductors that possess a high degree of symmetry. The solenoid and toroid are two classic examples where this law simplifies complex calculations.
1. Magnetic Field Due to a Long Solenoid (एक लंबी परिनालिका के कारण चुंबकीय क्षेत्र)
A solenoid is a long coil consisting of a large number of insulated turns of wire closely wound in the shape of a helix.
Key Results:
- Outside the Solenoid: The magnetic field ($\mathbf{B}$) is nearly zero ($\mathbf{B} \approx 0$). The field lines are confined almost entirely to the interior.
- Inside the Solenoid: The magnetic field is uniform (constant in magnitude and direction) and runs parallel to the axis of the solenoid.
Mathematical Derivation (using Ampère’s Law):
By choosing a rectangular Ampèrian loop (PQRS) spanning the solenoid:
$$\oint \mathbf{B} \cdot d\mathbf{l} = \int_{\text{P}}^{\text{Q}} \mathbf{B} \cdot d\mathbf{l} + \int_{\text{Q}}^{\text{R}} \mathbf{B} \cdot d\mathbf{l} + \int_{\text{R}}^{\text{S}} \mathbf{B} \cdot d\mathbf{l} + \int_{\text{S}}^{\text{P}} \mathbf{B} \cdot d\mathbf{l}$$
- $\mathbf{B} \cdot d\mathbf{l}$ along QR and SP is zero because $\mathbf{B}$ is perpendicular to $d\mathbf{l}$.
- $\mathbf{B} \cdot d\mathbf{l}$ along RS (outside the solenoid) is zero because $\mathbf{B} \approx 0$.
- Only the integral along PQ (length $l$) contributes.$$\int_{\text{P}}^{\text{Q}} \mathbf{B} \cdot d\mathbf{l} = B l$$
If $n$ is the number of turns per unit length, the current enclosed ($I_{\text{enclosed}}$) by the loop of length $l$ is $(nl)I$.
Applying Ampère’s Law:
$$B l = \mu_0 (n l I)$$
$$\mathbf{B} = \mu_0 n I$$
Hindi Translation: एक लंबी परिनालिका के भीतर चुंबकीय क्षेत्र एकरूप होता है और उसके अक्ष के समानांतर होता है। बाहर चुंबकीय क्षेत्र लगभग शून्य होता है। परिनालिका के भीतर क्षेत्र की तीव्रता $\mathbf{B} = \mu_0 n I$ होती है, जहाँ $n$ प्रति इकाई लंबाई में फेरों की संख्या है।
2. Magnetic Field Due to a Toroid (एक टोरोइड के कारण चुंबकीय क्षेत्र)
A toroid is a hollow circular ring on which a large number of insulated turns of a wire are closely wound. It can be seen as a solenoid bent into a circular shape.
Key Results:
The magnetic field is perfectly confined to the space inside the toroid’s coil.
- In the open space interior to the toroid (Region I): $\mathbf{B} = 0$.
- In the open space exterior to the toroid (Region III): $\mathbf{B} = 0$.
- Inside the Toroid (Region II): The field is uniform and concentric with the toroid.
Mathematical Derivation (using Ampère’s Law):
Choosing a circular Ampèrian loop (radius $r$) inside the toroid (Region II).
$$\oint \mathbf{B} \cdot d\mathbf{l} = B \cdot (2 \pi r)$$
If $N$ is the total number of turns, the total current enclosed ($I_{\text{enclosed}}$) is $N I$.
Applying Ampère’s Law:
$$B (2 \pi r) = \mu_0 N I$$
$$\mathbf{B} = \frac{\mu_0 N I}{2 \pi r}$$
- If we define $n$ as the number of turns per unit length ($n = \frac{N}{2\pi r}$), the expression becomes:$$\mathbf{B} = \mu_0 n I$$This is the same expression as that for a long solenoid, highlighting the close connection between the two devices.
Hindi Translation: एक टोरोइड (टोरोइडल परिनालिका) में, चुंबकीय क्षेत्र पूरी तरह से कॉइल के अंदर तक ही सीमित रहता है। टोरोइड के अंदर ($\mathbf{B}$) की तीव्रता $\mathbf{B} = \frac{\mu_0 N I}{2 \pi r}$ होती है, जहाँ $N$ कुल फेरों की संख्या और $r$ लूप की त्रिज्या है। यह सूत्र परिनालिका के सूत्र के समान ही है जब $n$ को प्रति इकाई लंबाई में फेरों की संख्या माना जाता है।
Ampère’s Law provides a straightforward way to calculate the magnetic fields for these symmetric geometries, saving significant effort compared to using the Biot-Savart Law.
Biot-Savart vs. Ampère’s Law: When to Use Which?
The choice between the Biot-Savart Law and Ampère’s Circuital Law depends entirely on the symmetry of the current distribution you’re analyzing. Both laws are fundamentally equivalent, but one is usually much easier to apply than the other.
1. 🔍 Use Biot-Savart Law (जब समरूपता न हो)
The Biot-Savart Law is the universal starting point for calculating the magnetic field ($\mathbf{B}$) from any current distribution. Use it when the geometry is complex or lacks symmetry.
| Reason for Use | Key Idea (मुख्य कारण) | Example (उदाहरण) |
| Lack of Symmetry | When the magnetic field $\mathbf{B}$ is not constant or not perpendicular/parallel to the path $d\mathbf{l}$ along any convenient closed loop. | Finding $\mathbf{B}$ at a point off-axis of a straight wire. |
| Defining the Field | When you need the field due to a small segment of current ($I d\mathbf{l}$), as it defines the field based on the segment. | Finding $\mathbf{B}$ at the center of a square loop or a circular arc. |
Hindi Translation: बायो-सावर्ट का नियम हर स्थिति में लागू होता है। इसका उपयोग तब करें जब धारा वितरण में समरूपता की कमी हो, जैसे कि एक वर्ग लूप के केंद्र में या किसी जटिल आकार के कारण चुंबकीय क्षेत्र की गणना करनी हो। इसमें पूरे पथ पर जटिल समाकलन की आवश्यकता होती है।
2. ⚡ Use Ampère’s Law (जब समरूपता हो)
Ampère’s Law acts as a shortcut and should be used whenever the current distribution possesses a high degree of symmetry, allowing you to choose a closed loop (Ampèrian loop) where $\mathbf{B}$ is constant and either parallel or perpendicular to the loop element $d\mathbf{l}$.
| Reason for Use | Key Idea (मुख्य कारण) | Example (उदाहरण) |
| High Symmetry | When the magnetic field $\mathbf{B}$ has constant magnitude and is tangential to the chosen closed loop, making $\oint \mathbf{B} \cdot d\mathbf{l} = B L$. | Magnetic field inside a long straight wire (both inside and outside). |
| Uniform Fields | For calculating the uniform field created inside devices specifically designed for symmetry. | Finding $\mathbf{B}$ inside a long solenoid or toroid. |
Hindi Translation: एम्पीयर के नियम का प्रयोग तब करें जब धारा वितरण में उच्च समरूपता हो। यह गॉस के नियम का चुंबकीय रूप है। यह लंबी सीधी तार, परिनालिका (solenoid), और टोरोइड के कारण $\mathbf{B}$ की गणना को सरल बनाता है क्योंकि आप एक ऐसा बंद लूप चुन सकते हैं जहाँ $\mathbf{B}$ का मान स्थिर हो।
Summary Table
| Law (नियम) | Nature (प्रकृति) | Best for (सर्वोत्तम) |
| Biot-Savart | Differential, Fundamental | Any geometry, but complex for symmetric cases. |
| Ampère’s | Integral, Simplified | High-symmetry geometries (Infinite wire, Solenoid, Toroid). |
Problem: Magnetic Field Inside a Long Straight Wire
Question: A long straight cylindrical wire of radius $R$ carries a steady current $I$ which is uniformly distributed throughout its cross-section. Find the magnetic field $\mathbf{B}$ at a point that is a distance $r$ from the center of the wire, where $r < R$ (i.e., inside the wire).
Step 1: Identify Symmetry and Choose Ampèrian Loop
The current is uniformly distributed and the wire is long and straight, indicating cylindrical symmetry. The magnetic field lines will be concentric circles around the wire’s axis.
- Ampèrian Loop: We choose a circular loop of radius $r$ ($r < R$) concentric with the wire.
- Along this loop, the magnetic field $\mathbf{B}$ is tangential (parallel to $d\mathbf{l}$) and has a constant magnitude.
Step 2: Calculate the Current Enclosed ($I_{\text{enclosed}}$)
Since the point is inside the wire ($r < R$), the Ampèrian loop encloses only a fraction of the total current $I$.
- Total Cross-sectional Area: $A_{\text{total}} = \pi R^2$
- Area Enclosed by Loop: $A_{\text{enclosed}} = \pi r^2$
- Current Density ($\text{J}$, constant): $J = \frac{\text{Total Current}}{\text{Total Area}} = \frac{I}{\pi R^2}$
- Current Enclosed:$$I_{\text{enclosed}} = J \times A_{\text{enclosed}} = \left(\frac{I}{\pi R^2}\right) \times (\pi r^2)$$$$I_{\text{enclosed}} = I \frac{r^2}{R^2}$$
Step 3: Apply Ampère’s Circuital Law
$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}$$
- Left-Hand Side (LHS): Since $\mathbf{B}$ and $d\mathbf{l}$ are parallel and $\mathbf{B}$ is constant on the loop:$$\oint \mathbf{B} \cdot d\mathbf{l} = B \oint dl = B (2 \pi r)$$(where $2\pi r$ is the circumference of the Ampèrian loop).
- Right-Hand Side (RHS): Substitute the expression for $I_{\text{enclosed}}$:$$\mu_0 I_{\text{enclosed}} = \mu_0 \left(I \frac{r^2}{R^2}\right)$$
Step 4: Equate LHS and RHS and Solve for B
$$B (2 \pi r) = \mu_0 I \frac{r^2}{R^2}$$
$$B = \frac{\mu_0 I r^2}{2 \pi r R^2}$$
Simplifying the term $\frac{r^2}{r}$:
$$\mathbf{B} = \frac{\mu_0 I r}{2 \pi R^2}$$
Conclusion (Inside the Wire, $r < R$)
The magnetic field $\mathbf{B}$ inside the wire increases linearly with the radial distance $r$ from the center.
$$\mathbf{B} \propto r$$
Problem: Magnetic Field Outside a Long Straight Wire
Question: A long straight cylindrical wire of radius $R$ carries a steady current $I$ which is uniformly distributed throughout its cross-section. Find the magnetic field $\mathbf{B}$ at a distance $r$ from the center of the wire, where $r > R$ (i.e., outside the wire).
Step 1: Identify Symmetry and Choose Ampèrian Loop
The current distribution still has cylindrical symmetry. The magnetic field lines are concentric circles.
- Ampèrian Loop: We choose a circular loop of radius $r$ ($r > R$) concentric with the wire.
- Along this loop, the magnetic field $\mathbf{B}$ is tangential and has a constant magnitude.
Step 2: Calculate the Current Enclosed ($I_{\text{enclosed}}$)
Since the Ampèrian loop now has a radius $r$ which is greater than the wire’s radius $R$, the loop encloses the entire current $I$ passing through the wire’s cross-section.
$$I_{\text{enclosed}} = I$$
Step 3: Apply Ampère’s Circuital Law
$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enclosed}}$$
- Left-Hand Side (LHS): Since $\mathbf{B}$ is constant and tangential to the loop of radius $r$:$$\oint \mathbf{B} \cdot d\mathbf{l} = B \oint dl = B (2 \pi r)$$
- Right-Hand Side (RHS): Substitute the total current $I$:$$\mu_0 I_{\text{enclosed}} = \mu_0 I$$
Step 4: Equate LHS and RHS and Solve for B
$$B (2 \pi r) = \mu_0 I$$
$$\mathbf{B} = \frac{\mu_0 I}{2 \pi r}$$
Conclusion (Outside the Wire, $r > R$)
The magnetic field $\mathbf{B}$ outside a long straight wire decreases inversely with the radial distance $r$ from the center. This is the same result you’d get if the wire were infinitesimally thin, carrying the current $I$.
$$\mathbf{B} \propto \frac{1}{r}$$
This result confirms that outside the wire, the field depends only on the total current enclosed, not the internal structure of the wire.
That’s a great transition! The Biot-Savart Law is essential for calculating fields when the symmetry required for Ampère’s Law is absent, like with a circular loop.
⭕ Problem: Magnetic Field at the Center of a Circular Current Loop
Question: A current $I$ flows through a circular wire loop of radius $R$. Determine the magnitude of the magnetic field $\mathbf{B}$ produced exactly at the center of the loop.
Step 1: Define the Current Element and Variables
We use the Biot-Savart Law in its differential form to find the contribution ($d\mathbf{B}$) from a small current element ($d\mathbf{l}$).
- Current Element ($d\mathbf{l}$): A small segment of the loop (tangential to the circle).
- Position Vector ($\mathbf{r}$): The vector pointing from the element $d\mathbf{l}$ to the center of the loop.
- Distance ($r$): The distance from $d\mathbf{l}$ to the center is constant and equal to the radius, $r = R$.
- Angle ($\theta$): The current element $d\mathbf{l}$ is always tangential, and the position vector $\mathbf{r}$ is always along the radius. Therefore, the angle between $d\mathbf{l}$ and $\mathbf{r}$ is always $\theta = 90^\circ$ ($\sin 90^\circ = 1$).
Step 2: Apply the Biot-Savart Law (Magnitude)
The magnitude of the differential magnetic field $d\mathbf{B}$ at the center due to $d\mathbf{l}$ is:
$$|d\mathbf{B}| = \frac{\mu_0}{4\pi} \frac{I d l \sin\theta}{r^2}$$
Substituting $r=R$ and $\sin\theta = 1$:
$$|d\mathbf{B}| = \frac{\mu_0}{4\pi} \frac{I d l}{R^2}$$
Step 3: Determine the Direction (Superposition)
Using the Right-Hand Rule for a circular loop (curl fingers in the direction of current, thumb points in the direction of $\mathbf{B}$):
- The direction of $d\mathbf{B}$ from every element $d\mathbf{l}$ is the same (perpendicular to the plane of the loop).
- Since all $d\mathbf{B}$ contributions are parallel, we can find the total field $\mathbf{B}$ by simply integrating the magnitudes.
Step 4: Integrate to Find the Total Field ($\mathbf{B}$)
The total magnetic field $\mathbf{B}$ is the integral of $d\mathbf{B}$ over the entire loop:
$$\mathbf{B} = \oint d\mathbf{B} = \oint \frac{\mu_0}{4\pi} \frac{I d l}{R^2}$$
Since $\mu_0, I, 4\pi,$ and $R^2$ are all constants, we pull them out of the integral:
$$\mathbf{B} = \frac{\mu_0 I}{4\pi R^2} \oint dl$$
The integral $\oint dl$ represents the sum of all small segments that make up the loop, which is the circumference of the loop, $2 \pi R$:
$$\oint dl = 2 \pi R$$
Step 5: Final Result
Substitute the circumference back into the equation:
$$\mathbf{B} = \frac{\mu_0 I}{4\pi R^2} (2 \pi R)$$
Simplifying the terms ($\frac{2\pi}{\pi} = 2$, $\frac{R}{R^2} = \frac{1}{R}$):
$$\mathbf{B} = \frac{\mu_0 I}{2 R}$$
Conclusion (Center of Circular Loop)
The magnitude of the magnetic field at the center of a single circular current loop is:
$$\mathbf{B} = \frac{\mu_0 I}{2 R}$$
If the loop has $N$ turns (like a tightly wound coil), the total field is multiplied by $N$:
$$\mathbf{B}_{\text{coil}} = \frac{\mu_0 N I}{2 R}$$
That’s the next level! Calculating the magnetic field along the axis of a circular current loop is a standard application that shows the power of the Biot-Savart Law and vector integration.
📐 Problem: Magnetic Field on the Axis of a Circular Loop
Question: A current $I$ flows through a circular wire loop of radius $R$. Determine the magnitude of the magnetic field $\mathbf{B}$ at a point $P$ located a distance $x$ from the center along the axis of the loop.
Step 1: Define Variables and Identify Cancellation
- Current Element ($d\mathbf{l}$): Choose a small element $d\mathbf{l}$ on the loop.
- Distance ($r’$): The distance from $d\mathbf{l}$ to point $P$ is constant for all elements on the loop:$$r’ = \sqrt{R^2 + x^2}$$
- Angle ($\theta$): The element $d\mathbf{l}$ is tangential to the circle and $\mathbf{r}’$ lies in the plane connecting $d\mathbf{l}$ and the axis. Since $d\mathbf{l}$ is perpendicular to the radial line on the loop and the radius is perpendicular to the axis, $d\mathbf{l}$ is perpendicular to $\mathbf{r}’$. Thus, $\mathbf{\theta = 90^\circ}$ ($\sin 90^\circ = 1$).
Vector Cancellation:
The differential field $d\mathbf{B}$ from $d\mathbf{l}$ is perpendicular to $\mathbf{r}’$ and lies in the plane defined by $\mathbf{r}’$ and $d\mathbf{l}$.
- When we resolve $d\mathbf{B}$ into components:
- The radial components ($d\mathbf{B}_{\perp}$) perpendicular to the axis cancel out for every pair of diametrically opposite current elements.
- Only the axial components ($d\mathbf{B}_{||}$) along the axis add up.
Step 2: Calculate the Axial Component ($d\mathbf{B}_{||}$)
We only need the component of $d\mathbf{B}$ along the axis. Let $\alpha$ be the angle between $d\mathbf{B}$ and the axis.
- $d\mathbf{B}_{||} = |d\mathbf{B}| \cos \alpha$
From the geometry, $\cos \alpha$ is the ratio of the radius ($R$) to the distance ($r’$):
$$\cos \alpha = \frac{R}{r’} = \frac{R}{\sqrt{R^2 + x^2}}$$
Step 3: Apply Biot-Savart Law and Integrate
Substitute the values for $r’$ and $\sin 90^\circ$ into the Biot-Savart Law:
$$|d\mathbf{B}| = \frac{\mu_0}{4\pi} \frac{I d l \sin 90^\circ}{(r’)^2} = \frac{\mu_0}{4\pi} \frac{I d l}{R^2 + x^2}$$
Now, calculate the axial component $d\mathbf{B}_{||}$:
$$d\mathbf{B}_{||} = |d\mathbf{B}| \cos \alpha = \left( \frac{\mu_0}{4\pi} \frac{I d l}{R^2 + x^2} \right) \left( \frac{R}{\sqrt{R^2 + x^2}} \right)$$
$$d\mathbf{B}_{||} = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} d l$$
To find the total magnetic field $\mathbf{B}$, we integrate $d\mathbf{B}_{||}$ over the entire loop ($\oint dl$):
$$\mathbf{B} = \oint d\mathbf{B}_{||} = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \oint dl$$
Since $\oint dl$ is the circumference of the loop, $\mathbf{2\pi R}$:
$$\mathbf{B} = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} (2 \pi R)$$
Step 4: Simplify the Final Expression
$$\mathbf{B} = \frac{\mu_0 I (2 \pi R^2)}{4\pi (R^2 + x^2)^{3/2}}$$
Canceling the $2\pi$ terms and simplifying:
$$\mathbf{B} = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}$$
Key Takeaways and Special Cases
- General Result (Axis): The magnetic field at a distance $x$ from the center along the axis is:$$\mathbf{B} = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}$$
- Field at the Center ($x=0$): If we set $x=0$, the formula simplifies to:$$\mathbf{B}_{\text{center}} = \frac{\mu_0 I R^2}{2 (R^2 + 0)^{3/2}} = \frac{\mu_0 I R^2}{2 R^3} = \frac{\mu_0 I}{2R}$$This correctly matches the result we derived earlier using a simplified application of the Biot-Savart Law!
- Field Far Away ($x \gg R$): If the point is very far away, we can ignore $R^2$ in the denominator:$$\mathbf{B}_{\text{far}} \approx \frac{\mu_0 I R^2}{2 (x^2)^{3/2}} = \frac{\mu_0 I R^2}{2 x^3}$$The field strength decreases rapidly, dropping off as the inverse cube ($1/x^3$) of the distance.
⚡ Magnetic Force on a Moving Charge (Lorentz Force)
A magnetic field ($\mathbf{B}$) doesn’t exert a force on a stationary charge, but it exerts a force on a charge that is moving within the field. This force is a component of the total Lorentz Force (which includes the electric force).
1. The Mathematical Formula
The magnetic force ($\mathbf{F}_m$) acting on a point charge $q$ moving with velocity $\mathbf{v}$ in a uniform magnetic field $\mathbf{B}$ is given by the vector cross product:
$$\mathbf{F}_m = q (\mathbf{v} \times \mathbf{B})$$
The magnitude of the force is:
$$F_m = q v B \sin\theta$$
Where:
- $q$ is the magnitude of the charge (C).
- $v$ is the speed of the charge ($\text{m/s}$).
- $B$ is the magnitude of the magnetic field (Tesla, T).
- $\theta$ is the angle between the velocity vector ($\mathbf{v}$) and the magnetic field vector ($\mathbf{B}$).
2. Direction of the Force
The magnetic force $\mathbf{F}_m$ is always perpendicular to both the velocity ($\mathbf{v}$) of the charge and the magnetic field ($\mathbf{B}$).
The direction is determined using Fleming’s Left-Hand Rule or the Right-Hand Rule for vector cross products:
- For a positive charge ($q > 0$): Use the standard Right-Hand Rule (Fingers point in $\mathbf{v}$, curl towards $\mathbf{B}$, thumb points in $\mathbf{F}_m$).
- For a negative charge ($q < 0$): The force is in the opposite direction to that given by the Right-Hand Rule.
3. Key Properties and Special Cases
| Case | Angle (θ) | Magnitude (Fm) | Force and Motion |
| Maximum Force | $\theta = 90^\circ$ ($\mathbf{v} \perp \mathbf{B}$) | $F_{max} = qvB$ | The charge moves in a circular path. |
| Zero Force | $\theta = 0^\circ$ or $180^\circ$ ($\mathbf{v} \parallel \mathbf{B}$) | $F_{min} = 0$ | The charge continues to move in a straight line at constant speed. |
| Helical Path | $0^\circ < \theta < 90^\circ$ | $F_m = qvB \sin\theta$ | The charge moves in a helical (spiral) path. |
Crucial Point: Since the force $\mathbf{F}_m$ is always perpendicular to the velocity $\mathbf{v}$, the magnetic force never does any work on the charge ($W = \mathbf{F} \cdot d\mathbf{s} = 0$) and therefore does not change the kinetic energy or speed of the charged particle. It only changes the direction of motion.
⚙️ Problem: Radius of the Circular Path
Question: A charged particle of mass $m$ and charge $q$ enters a uniform magnetic field $\mathbf{B}$ with a velocity $\mathbf{v}$. If the velocity $\mathbf{v}$ is perpendicular to the magnetic field $\mathbf{B}$ ($\theta = 90^\circ$), derive the expression for the radius ($r$) of the resulting circular path.
Step 1: Force and Motion
When the velocity $\mathbf{v}$ is perpendicular to the field $\mathbf{B}$, the magnetic force $\mathbf{F}_m$ is always perpendicular to the velocity. This is the condition required for uniform circular motion.
- The magnetic force acts as the necessary centripetal force ($F_c$) required to keep the particle moving in a circle.
$$\mathbf{F}_m = \mathbf{F}_c$$
Step 2: Express the Forces
- Magnetic Force ($\mathbf{F}_m$):$$F_m = q v B \sin 90^\circ = q v B$$
- Centripetal Force ($F_c$):$$F_c = \frac{m v^2}{r}$$
Step 3: Equate the Forces and Solve for Radius ($r$)
Set the two forces equal to each other:
$$q v B = \frac{m v^2}{r}$$
Now, solve for the radius ($r$):
- Cancel one $v$ from both sides:$$q B = \frac{m v}{r}$$
- Rearrange the equation to isolate $r$:$$r = \frac{m v}{q B}$$
Conclusion: Radius of Circular Path
The radius of the circular path taken by a charged particle moving perpendicular to a uniform magnetic field is:
$$r = \frac{m v}{q B}$$
- This shows that the radius is directly proportional to the momentum ($p = mv$) of the particle and inversely proportional to the charge ($q$) and the magnetic field strength ($B$).
Additional Key Parameters
Based on the derived radius, we can find the frequency and time period of the motion:
- Angular Frequency ($\omega$):$$\omega = \frac{v}{r} = \frac{v}{(mv/qB)} = \frac{q B}{m}$$
- Time Period ($T$): The time taken for one full revolution.$$T = \frac{2\pi r}{v} = \frac{2\pi (mv/qB)}{v}$$$$T = \frac{2 \pi m}{q B}$$
- Cyclotron Frequency ($f$): (Frequency of revolution)$$f = \frac{1}{T} = \frac{q B}{2 \pi m}$$
Notice that the Time Period ($T$) and Frequency ($f$) are independent of the speed ($v$) and the radius ($r$). This critical property is utilized in devices like the Cyclotron.