The magnitude of the magnetic field of this wire is about 5.67 × 10⁻⁶ T
[tex]\texttt{ }[/tex]
Further explanation
Let's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:
[tex]\boxed {B = \mu_o \frac{I}{2 \pi d} } [/tex]
B = magnetic field strength from current carrying wire (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
d = distance (m)
[tex]\texttt{ }[/tex]
[tex]\boxed {B = \mu_o \frac{I N}{L} } [/tex]
B = magnetic field strength at the center of the solenoid (T)
μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)
I = current (A)
N = number of turns
L = length of solenoid (m)
Let's tackle the problem now !
[tex]\texttt{ }[/tex]
Given:
Number of Electrons per second = N/t = 8.15 × 10¹⁸ electrons/second
Distance = d = 4.60 cm = 0.046 m
Permeability of free space = μo = 4π × 10⁻⁷ T.m/A
Charge of Electron = e = 1.6 × 10⁻¹⁹ C
Asked:
Wire's Magnetic Field Strength = B = ?
Solution:
We will use this folllowing formula to solve the problem:
[tex]B = \mu_o \frac{I}{2 \pi d}[/tex]
[tex]B = \mu_o \frac{Q}{2 \pi d t}[/tex]
[tex]B = \mu_o \frac{Ne}{2 \pi d t}[/tex]
[tex]B = 4\pi \times 10^{-7} \times \frac{8.15 \times 10^{18} \times 1.6 \times 10^{-19}}{2 \pi \times 0.046}[/tex]
[tex]\boxed{B \approx 5.67 \times 10^{-6} \texttt{ T}}[/tex]
[tex]\texttt{ }[/tex]
Learn more
- Temporary and Permanent Magnet : https://brainly.com/question/9966993
- The three resistors : https://brainly.com/question/9503202
- A series circuit : https://brainly.com/question/1518810
- Compare and contrast a series and parallel circuit : https://brainly.com/question/539204
[tex]\texttt{ }[/tex]
Answer details
Grade: High School
Subject: Physics
Chapter: Magnetic Field
