A long straight wire carries a current of $10 ext{ A}$. Calculate the magnitude of the magnetic field produced at a perpendicular distance of $5 ext{ cm}$ from the wire. Given the permeability of free space $\mu0 = 4\pi imes 10^{-7} ext{ T}\cdot ext{m/A}$. Also, state the rule used to find the direction of this magnetic field and explain it briefly.

Q: A long straight wire carries a current of $10\text{ A}$. Calculate the magnitude of the magnetic field produced at a perpendicular distance of $5\text{ cm}$ from the wire. Given the permeability of free space $\mu0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$. Also, state the rule used to find the direction of this magnetic field and explain it briefly.

Given Data: Current ($I$) = $10\text{ A}$ Distance ($r$) = $5\text{ cm} = 5 \times 10^{-2} \text{ m}$ Permeability of free space ($\mu0$) = $4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$ Formula:\nThe magnetic field ($B$) at a perpendicular distance $r$ from a long straight current-carrying wire is given by: $$B = \frac{\mu0 I}{2\pi r}$| Calculation:\nSubstitute the given values into the formula: $$B = \frac{(4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \times (10 \text{ A})}{2\pi \times (5 \times 10^{-2} \text{ m})}|$$ \nSimplify the expression: $$B = \frac{4\pi \times 10^{-6}}{10\pi \times 10^{-2}}$$ $$B = \frac{4\pi \times 10^{-6}}{\pi \times 10^{-1}}$$ $$B = 4 \times 10^{-6 - (-1)} \text{ T}$$ $$B = 4 \times 10^{-5} \text{ Tesla} (\text{or } 40 \mu\text{T})$| Direction of Magnetic Field:\nTo find the direction of the magnetic field produced by a straight current-carrying wire, we use Maxwell's Right-Hand Thumb Rule (also known as the Right-Handed Screw Rule). Statement of Maxwell's Right-Hand Thumb Rule:\nImagine that you are holding a current-carrying straight wire in your right hand such that your thumb points in the direction of the current. Then your fingers will wrap around the wire in the direction of the field lines of the magnetic field.

Question

A long straight wire carries a current of $10	ext{ A}$. Calculate the magnitude of the magnetic field produced at a perpendicular distance of $5	ext{ cm}$ from the wire. Given the permeability of free space $\mu0 = 4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}$. Also, state the rule used to find the direction of this magnetic field and explain it briefly.

Accepted Answer

Given Data: Current ($I$) = $10	ext{ A}$ Distance ($r$) = $5	ext{ cm} = 5 	imes 10^{-2} 	ext{ m}$ Permeability of free space ($\mu0$) = $4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}$ Formula:
The magnetic field ($B$) at a perpendicular distance $r$ from a long straight current-carrying wire is given by: $$B = \frac{\mu0 I}{2\pi r}$| Calculation:
Substitute the given values into the formula: $$B = \frac{(4\pi 	imes 10^{-7} 	ext{ T}\cdot	ext{m/A}) 	imes (10 	ext{ A})}{2\pi 	imes (5 	imes 10^{-2} 	ext{ m})}|$$ 
Simplify the expression: $$B = \frac{4\pi 	imes 10^{-6}}{10\pi 	imes 10^{-2}}$$ $$B = \frac{4\pi 	imes 10^{-6}}{\pi 	imes 10^{-1}}$$ $$B = 4 	imes 10^{-6 - (-1)} 	ext{ T}$$ $$B = 4 	imes 10^{-5} 	ext{ Tesla} (	ext{or } 40 \mu	ext{T})$| Direction of Magnetic Field:
To find the direction of the magnetic field produced by a straight current-carrying wire, we use Maxwell's Right-Hand Thumb Rule (also known as the Right-Handed Screw Rule). Statement of Maxwell's Right-Hand Thumb Rule:
Imagine that you are holding a current-carrying straight wire in your right hand such that your thumb points in the direction of the current. Then your fingers will wrap around the wire in the direction of the field lines of the magnetic field.