The magnitude of the magnetic field of this wire is about 5.67 × 10⁻⁶ T
<h3>Further explanation</h3>
<em>Let's recall </em><em>magnetic field strength</em><em> from current carrying wire and from center of the solenoid as follows:</em>
<em>B = magnetic field strength from current carrying wire (T)</em>
<em>μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)</em>
<em>I = current (A)</em>
<em>d = distance (m)</em>
<em>B = magnetic field strength at the center of the solenoid (T)</em>
<em>μo = permeability of free space = 4π × 10⁻⁷ (Tm/A)</em>
<em>I = current (A)</em>
<em>N = number of turns</em>
<em>L = length of solenoid (m)</em>
Let's tackle the problem now !
<u>Given:</u>
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
<u>Asked:</u>
Wire's Magnetic Field Strength = B = ?
<u>Solution:</u>
<em>We will use this folllowing formula to solve the problem:</em>
<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Magnetic Field