Question

A very long, straight horizontal wire carries a current such that 8.15×1018 electrons per second pass any given point going from west to east. part a what is the magnitude of the magnetic field this wire produces at a point 4.60 cm directly above it

Answer 1

The magnitude of the magnetic field of this wire is about 5.67 × 10⁻⁶ T

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Further explanation

Let's recall magnetic field strength from current carrying wire and from center of the solenoid as follows:

$\boxed {B = \mu_o \frac{I}{2 \pi d} }$

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)

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$\boxed {B = \mu_o \frac{I N}{L} }$

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 !

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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:

$B = \mu_o \frac{I}{2 \pi d}$

$B = \mu_o \frac{Q}{2 \pi d t}$

$B = \mu_o \frac{Ne}{2 \pi d t}$

$B = 4\pi \times 10^{-7} \times \frac{8.15 \times 10^{18} \times 1.6 \times 10^{-19}}{2 \pi \times 0.046}$

$\boxed{B \approx 5.67 \times 10^{-6} \texttt{ T}}$

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Magnetic Field

Answer 2

Magnetic field 'B' at a distance 'r' from an substantially large conductor carrying current 'i' = (2x10^ -7)('i' ) / r 

Magnetic field 'B' beyond the wire= (2x10^ -7)(8.15x10^18x1.6x10^ - 19 ) /0.046 =5.7 x10^ -12 tesla 

As electrons move from west to east, the conventional current is from east to west. 

By means of Maxwell's right handed corkscrew rule, the way of magnetic field is from south to north.