Respuesta :
Answer:
Magnetic field, [tex]B=8.69\times 10^{-5}\ T[/tex]
Explanation:
Given that,
Charge, q = 15 C
Time taken, [tex]t=1.5\times 10^{-3}\ s[/tex]
Distance from the blot, d = 23 m
Let I is the current. It is equal to the charge per unit time. It can be calculated as :
[tex]I=\dfrac{q}{t}[/tex]
[tex]I=\dfrac{15}{1.5\times 10^{-3}}[/tex]
I = 10000 A
The magnetic field at a distance d from the bolt is given by :
[tex]B=\dfrac{\mu_oI}{2\pi d}[/tex]
[tex]B=\dfrac{4\pi \times 10^{-7}\times 10000}{2\pi \times 23}[/tex]
[tex]B=8.69\times 10^{-5}\ T[/tex]
So, the magnetic field at a distance of 23 m from the bolt is [tex]8.69\times 10^{-5}\ T[/tex]. Hence, this is the required solution.
The magnitude of the magnetic field ( β ) is : 8.69 * 10⁻⁵ T
Given data :
charge ( q ) = 15 C
time = 1.5 * 10⁻³ s
distance from bolt = 23 m
Determine the magnitude of the magnetic field
First step : Calculate the value of current
I = q / T
where : q = charge , T [ time
I = 15 / ( 1.5 * 10⁻³ )
= 10000 A .
Final step : calculate the magnitude of the magnetic field at 23 m form the bolt
β = u₀I / 2[tex]\pi d[/tex]
= [tex]\frac{4\pi * 10^{-7} *10000 }{2\pi * 23}[/tex]
= 8.69 * 10⁻⁵ T
Hence we can conclude that The magnitude of the magnetic field ( β ) is : 8.69 * 10⁻⁵ T
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