Problem 30
Question
Consider a coaxial cable which consists of a wire of radius \(a\) and outer
cylindrical shell of inner and outer radii \(b\) and \(c\) respectively. The inner
wire carries a current \(i\) and outershell carries an equal and opposite
current. The magnetic field at a distance \(x\) from axis from \(x
Step-by-Step Solution
Verified Answer
The magnetic field at distance \(x
1Step 1: Define the Problem
We are dealing with a coaxial cable where a wire of radius \(a\) carries a current \(i\), and an outer cylindrical shell carries an equal and opposite current. The goal is to find the magnetic field at a distance \(x\) from the axis when \(x < a\).
2Step 2: Use Ampere's Law
Ampere's Law relates the magnetic field around a closed loop to the current passing through the loop. It is given by: \[ \oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{enc} \] where \(I_{enc}\) is the current enclosed by the loop. For a circle of radius \(x\) within the wire (\(x < a\)), only a portion of \(i\) is enclosed.
3Step 3: Calculate Enclosed Current
For \(x < a\), the current enclosed \(I_{enc}\) is proportional to the area of the circle of radius \(x\) relative to the area of the wire \(a\). Thus, \[ I_{enc} = i \cdot \left(\frac{\pi x^{2}}{\pi a^{2}}\right) = i \cdot \left(\frac{x^{2}}{a^{2}}\right) \] This shows that only a fraction of the total current is enclosed for \(x < a\).
4Step 4: Apply Ampere's Law
Now, substitute \(I_{enc}\) into Ampere's Law equation: \[ B \cdot (2\pi x) = \mu_{0} \cdot \left( i \cdot \frac{x^{2}}{a^{2}} \right) \] Solving for \(B\), we get: \[ B = \frac{\mu_{0} i x}{2 \pi a^{2}} \]
5Step 5: Conclusion
The magnetic field at a distance \(x\) from the axis where \(x < a\) is \(\frac{\mu_{0} i x}{2 \pi a^{2}}\). Therefore, option (d) is the correct answer.
Key Concepts
Coaxial CableMagnetic Field CalculationCurrent Distribution
Coaxial Cable
A coaxial cable is a type of electrical cable where a central conductor is surrounded by a cylindrical insulating layer, which is then enveloped by a tubular conducting shield. This structure is critical in ensuring the cable supports and transmits electrical signals with minimal interference. Coaxial cables are commonly used in television, internet, and radio frequency applications for their ability to reduce electromagnetic interference (EMI).
- **Inner Conductor**: Typically made of chosen metals such as copper or aluminum, it carries the electric signal.
- **Insulation**: Separates and insulates the inner conductor from the outer shield.
- **Outer Shield**: Often a braided, tubular conductor that blocks external signals that could distort the transmitted signal.
- **Protective Insulation**: An outer synthetic layer wraps the entire cable to protect the inner components. In practices related to physics, studying coaxial cables also illuminate the concepts of electromagnetism, such as magnetic fields created by current flows within the cable.
- **Inner Conductor**: Typically made of chosen metals such as copper or aluminum, it carries the electric signal.
- **Insulation**: Separates and insulates the inner conductor from the outer shield.
- **Outer Shield**: Often a braided, tubular conductor that blocks external signals that could distort the transmitted signal.
- **Protective Insulation**: An outer synthetic layer wraps the entire cable to protect the inner components. In practices related to physics, studying coaxial cables also illuminate the concepts of electromagnetism, such as magnetic fields created by current flows within the cable.
Magnetic Field Calculation
In electromagnetism, the calculation of a magnetic field involves the application of Ampere's Law. This is especially relevant in systems like the coaxial cable where currents flow in opposing directions in the inner conductor and the outer shell. Ampere's Law states:\[ \oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{enc} \]where \(\oint \vec{B} \cdot d\vec{l}\) is the path integral of the magnetic field and \(I_{enc}\) is the current enclosed by the path.
To find the magnetic field at a point inside the cable (distance \(x\) from the axis, where \(x < a\)), only the portion of the current within radius \(x\) contributes to the magnetic field. The formula becomes: - The magnetic field strength \(B\) is directly proportional to the current and the distance \(x\), while being inversely proportional to the square of the radius of the inner wire.- Resultantly, the strength formula \(B = \frac{\mu_{0} i x}{2 \pi a^{2}}\) highlights the linear relationship between the magnetic field and distance at positions within the inner conductor.
To find the magnetic field at a point inside the cable (distance \(x\) from the axis, where \(x < a\)), only the portion of the current within radius \(x\) contributes to the magnetic field. The formula becomes: - The magnetic field strength \(B\) is directly proportional to the current and the distance \(x\), while being inversely proportional to the square of the radius of the inner wire.- Resultantly, the strength formula \(B = \frac{\mu_{0} i x}{2 \pi a^{2}}\) highlights the linear relationship between the magnetic field and distance at positions within the inner conductor.
Current Distribution
Understanding current distribution is key to analyzing magnetic fields in conductors and especially in coaxial cables. In a coaxial cable, current flows are distributed as follows:
- **Inner Conductor**: Carries a uniform current, denoted as \(i\), down its length.- **Outer Shell**: Carries an equal but opposite current, creating a cancellation effect at any point outside the cable in terms of magnetic field effects.
Inside the inner wire, current is distributed evenly throughout the cross-section of the wire. Hence, for any location \(x < a\) within this wire, the enclosed current is proportional to the area of the cross-section at that radius, calculated as:\[ I_{enc} = i \cdot \left(\frac{x^{2}}{a^{2}}\right) \]This ensures that when analyzing magnetic field effects within a coaxial cable, only the portion of the current that physically surrounds a point up to \(x\) affects magnetic field calculations. The positive relation between the enclosed area and current exemplifies the fractional current flow contributing within specific zones inside the conductor.
- **Inner Conductor**: Carries a uniform current, denoted as \(i\), down its length.- **Outer Shell**: Carries an equal but opposite current, creating a cancellation effect at any point outside the cable in terms of magnetic field effects.
Inside the inner wire, current is distributed evenly throughout the cross-section of the wire. Hence, for any location \(x < a\) within this wire, the enclosed current is proportional to the area of the cross-section at that radius, calculated as:\[ I_{enc} = i \cdot \left(\frac{x^{2}}{a^{2}}\right) \]This ensures that when analyzing magnetic field effects within a coaxial cable, only the portion of the current that physically surrounds a point up to \(x\) affects magnetic field calculations. The positive relation between the enclosed area and current exemplifies the fractional current flow contributing within specific zones inside the conductor.
Other exercises in this chapter
Problem 29
A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity \(\mathbf{
View solution Problem 30
A particle of charge \(q\) and mass \(m\) starts moving from the origin under the action of an electric field, \(E=E_{0}\) î and \(B=B_{0}\) î with a velocity,
View solution Problem 31
A proton, a deutron and an \(\alpha\)-particle with the same kinetic energy enter a region of uniform magnetic field, moving at right angle to \(B\). What is th
View solution Problem 32
A proton of mass \(1.67 \times 10^{-27} \mathrm{~kg}\) and charge \(1.6 \times 10^{-19} \mathrm{C}\) is projected with a speed of \(2 \times 10^{6} \mathrm{~ms}
View solution