Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- Magnetic Field due to a Current
Concepts Of Physics MCQ Edition [Volume 2] · Physics · Magnetic Field due to a Current
Determine the maximum magnitude of the magnetic field B produced by...
Determine the maximum magnitude of the magnetic field produced by a copper wire having a diameter of and carrying a current of .
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
Given:
Diameter of the wire,
Radius of the wire,
Current,
The magnetic field produced by a current-carrying wire is maximum at its surface.
Using Ampere's law, the maximum magnetic field is given by:
B_{\text\max} = \dfrac{\mu_0 I}{2 \pi R}
Substituting the given values:
B_{\text\max} = \dfrac{4 \pi \times 10^{-7} \times 20}{2 \pi \times 0.8 \times 10^{-3}}
B_{\text\max} = \dfrac{2 \times 10^{-7} \times 20}{0.8 \times 10^{-3}}
B_{\text\max} = \dfrac{40 \times 10^{-7}}{8 \times 10^{-4}}
B_{\text\max} = 5.0 \times 10^{-3} \text{ T}
Diameter of the wire,
Radius of the wire,
Current,
The magnetic field produced by a current-carrying wire is maximum at its surface.
Using Ampere's law, the maximum magnetic field is given by:
B_{\text\max} = \dfrac{\mu_0 I}{2 \pi R}
Substituting the given values:
B_{\text\max} = \dfrac{4 \pi \times 10^{-7} \times 20}{2 \pi \times 0.8 \times 10^{-3}}
B_{\text\max} = \dfrac{2 \times 10^{-7} \times 20}{0.8 \times 10^{-3}}
B_{\text\max} = \dfrac{40 \times 10^{-7}}{8 \times 10^{-4}}
B_{\text\max} = 5.0 \times 10^{-3} \text{ T}