Magnetic Effects Of Current And Magnetism

• Electrons revolve in fixed orbits without emitting energy.
• Only certain discrete orbits are allowed where angular momentum is quantized.
• Electrons emit or absorb energy when transitioning between orbits.

According to Bohr's quantization condition:

For ground state :

Total energy of an electron in the ground state:

For :

Step 1: Biot-Savart Law.
The Biot-Savart Law gives the magnetic field produced at a point due to a small current element as: where:
- is the permeability of free space ( ),
- is the current,
- is the vector length of the current element,
- is the unit vector from the current element to the point where the magnetic field is calculated,
- is the distance from the current element to the point.

Step 2: Magnetic Field Due to a Long Straight Conductor.
For an infinitely long, straight conductor carrying a current , we can integrate the Biot-Savart law along the length of the conductor. The magnetic field at a distance from the conductor is given by: This expression is derived by integrating the Biot-Savart law for an infinite length conductor. The direction of the magnetic field follows the right-hand rule, meaning the magnetic field circulates around the wire in concentric circles.

Final Answer: The magnetic field produced by an infinitely long current-carrying conductor is given by: where is the distance from the wire, and the direction is given by the right-hand rule.

Step 1: Statement of Ampere's Circuital Law:
Ampere's law states that the line integral of the magnetic field around any closed loop (called an Amperian loop) is equal to times the total current passing through the area enclosed by the loop.

Step 2: Derivation for a Long Solenoid:
Consider a long solenoid with turns per unit length, carrying a current .
\begin{enumerate} \item Symmetry and Magnetic Field: For an ideal long solenoid, the magnetic field inside is strong, uniform, and directed along the axis of the solenoid. The magnetic field outside is negligibly weak (approximately zero).
\item Amperian Loop: To apply Ampere's law, we choose a rectangular Amperian loop `pqrs` of length , as shown in the cross-sectional diagram. The side `pq` is inside the solenoid, parallel to the axis, while the side `rs` is outside.
\begin{center} \begin{tikzpicture} % Solenoid turns (cross-section) \foreach \x in {0,1,2,3,4,5,6} { \node at (\x, 0.5) { }; % Current into page \node at (\x, -0.5) { }; % Current out of page } \draw[gray] (-0.5, 0.8) -- (6.5, 0.8); \draw[gray] (-0.5, -0.8) -- (6.5, -0.8); % Magnetic field lines \foreach \y in {-0.3,0,0.3} { \draw[->, blue, thick] (-0.5, \y) -- (6.5, \y); } \node at (3, 0) [above, blue] { }; \node at (3, 1.2) [above, blue] { }; % Amperian loop \draw[red, thick, ->] (1, -0.2) -- (5, -0.2) node[midway, below] { }; \draw[red, thick, ->] (5, -0.2) -- (5, 1.2) node[midway, right] { }; \draw[red, thick, ->] (5, 1.2) -- (1, 1.2) node[midway, above] { }; \draw[red, thick, ->] (1, 1.2) -- (1, -0.2) node[midway, left] { }; \node at (0.8, -0.2) [below left] { }; \node at (5.2, -0.2) [below right] { }; \node at (5.2, 1.2) [above right] { }; \node at (0.8, 1.2) [above left] { }; \draw[<->] (1,-1) -- (5,-1) node[midway, below] { }; \end{tikzpicture} \end{center} \item Evaluate the Line Integral: We evaluate for the loop `pqrs`. \begin{itemize} \item Along pq (inside): is parallel to . So, . \item Along qr and sp (perpendicular): is perpendicular to . So, . \item Along rs (outside): The magnetic field outside is zero ( ). So, . \end{itemize} The total value of the line integral is: . \item Calculate Enclosed Current ( ): The number of turns within the loop of length is . Since each turn carries current , the total current enclosed by the loop is: \item Apply Ampere's Law: \item Solve for B: Canceling from both sides gives the magnetic field inside the solenoid: \end{enumerate}

Step 1: Understanding the Concept:
A long, straight wire carrying a current produces a magnetic field in the form of concentric circles around the wire. The magnitude of this field can be calculated using Ampere's Law or the Biot-Savart Law.

Step 2: Key Formula or Approach:
For an infinitely long, straight wire carrying a current , the magnitude of the magnetic field at a perpendicular distance from the wire is given by: where is the permeability of free space, with a value of .

Step 3: Detailed Explanation:
First, identify the given values and convert them to SI units: \begin{itemize} \item Current, A.
\item The wire is along the Z-axis.
\item The point is P(10 cm, 0, 0).
\item The perpendicular distance of the point P from the Z-axis is its x-coordinate, which is 10 cm.
\item Convert the distance to meters: .
\end{itemize} Now, substitute these values into the formula for the magnetic field: Cancel out from the numerator and denominator:

Step 4: Final Answer:
The intensity of the magnetic field at the point (10 cm, 0, 0) is Tesla.
(Note: To find the direction, use the right-hand thumb rule. If the thumb points along the current (+Z direction), the curled fingers at the point (+X axis) will point in the +Y direction. So, ).

Step 1: Understanding the Concept:
Two long, parallel wires carrying currents exert a magnetic force on each other. The magnetic field produced by one wire interacts with the current in the other wire. When the currents are in the same direction, the force is attractive. The magnitude of this force per unit length is given by a standard formula.

Step 2: Key Formula or Approach:
The force per unit length ( ) between two parallel wires carrying currents and and separated by a distance is: In this problem, the currents are the same, so . The formula becomes:

Step 3: Detailed Explanation (Calculation):
We are given the following values: \begin{itemize} \item Force per unit length, N/m. \item Distance between wires, m. \item Permeability of free space, T·m/A. \end{itemize} Substitute these values into the formula and solve for : Cancel from the numerator and denominator: Rearrange to solve for : Take the square root to find the current :

Step 4: Final Answer:
The value of the current flowing in each wire is 30 A.

Step 1: Understanding the Concept:
An electron revolving in a circular orbit is equivalent to a tiny current loop. According to Ampere's law, any current loop produces a magnetic field and has an associated magnetic dipole moment. The magnitude of this magnetic moment depends on the equivalent current and the area of the loop.

Step 2: Key Formula or Approach:
1. The magnetic dipole moment ( ) of a current loop is given by , where is the current and is the area of the loop.
2. The equivalent current produced by an electron (charge ) revolving with a time period is .
3. The time period of revolution is the circumference of the orbit divided by the speed: .
Combining these formulas, we get an expression for the magnetic moment:

Step 3: Detailed Explanation:
We are given the following values: \begin{itemize} \item Speed of the electron, m/s. \item Radius of the orbit, m. \item Charge of an electron, C. \end{itemize} Now, we substitute these values into the derived formula for magnetic moment:

Step 4: Final Answer:
The magnetic moment of the rotating electron is A·m .

Step 1: Understanding the Concept:
The magnetic field produced by a current-carrying wire can be calculated using the Biot-Savart Law. This law relates the magnetic field to the current element, its length, and the position vector to the point of interest.

Step 2: Key Formula or Approach:
The Biot-Savart Law is given by: Here, is the vector representing the current element and is the unit vector pointing from the current element to the point O where the field is being calculated.

Step 3: Detailed Explanation:
For both straight portions of the wire, the point O lies on the axis of the wire. This means for any current element on the straight wires, the position vector pointing to O is either parallel (angle of 0 ) or anti-parallel (angle of 180 ) to .
The cross product has a magnitude of .
Since or , .
Therefore, the cross product is zero, and the magnetic field contribution from both straight portions at point O is zero.
(ii) Magnetic Field due to Semi-circular Portion

Step 1: Understanding the Concept:
A current flowing through a circular arc produces a magnetic field at the center of the arc. The magnitude of this field is proportional to the current and the angle subtended by the arc at the center.

Step 2: Key Formula or Approach:
The magnetic field at the center of a full circular loop of radius R carrying current i is: A semi-circle is half of a full circle, so it subtends an angle of radians at the center. The field it produces will be half of that of a full circle.

Step 3: Detailed Explanation:
The magnetic field due to the semi-circular portion of radius R is: Direction: To find the direction of the magnetic field, we use the Right-Hand Grip Rule. If we curl the fingers of our right hand in the direction of the current in the semi-circle (counter-clockwise in the diagram provided, although the arrows suggest a left-to-right flow), the thumb points in the direction of the magnetic field. For the current shown flowing from left to right, the thumb points into the plane of the page.

Step 4: Final Answer:
(i) The magnetic field at O due to each straight portion is zero.
(ii) The value of the magnetic field at O due to the semi-circle of radius R is , directed into the plane of the page.