Respuesta :
By using the formula:
Tau=(I)(A)(B)Sinθ
Get the area first of the circle:
A= πr^2
A= π(0.08m)^2
A= 0.02011 m^2
Then solve,
B(earth)= 5x10^-5 T
θ = 0, since the angle of the wire is oriented for maximum torque in the earth's field. Which means the angle is 0.
Reorder the formula to solve for the charge (I):
I = (tau)/(A)(B)Sinθ
I = (tau)/(A)(B)
I = (1.0x10^-3) / (0.02011)(5x10^-5)
I = 994 A
Tau=(I)(A)(B)Sinθ
Get the area first of the circle:
A= πr^2
A= π(0.08m)^2
A= 0.02011 m^2
Then solve,
B(earth)= 5x10^-5 T
θ = 0, since the angle of the wire is oriented for maximum torque in the earth's field. Which means the angle is 0.
Reorder the formula to solve for the charge (I):
I = (tau)/(A)(B)Sinθ
I = (tau)/(A)(B)
I = (1.0x10^-3) / (0.02011)(5x10^-5)
I = 994 A
The current required in the loop to experience the given torque is [tex]\boxed{6.366\times{{10}^2}\,{\text{A}}}[/tex] or [tex]\boxed{636.6\,{\text{A}}}[/tex] .
Further Explanation:
Given:
The diameter of the circular loop is [tex]20\,{\text{cm}}[/tex] .
The torque experienced by the circular loop is [tex]1.0\times{10^{-3}}\,{\text{N}}\cdot{\text{m}}[/tex] .
Concept:
Since the circular loop is kept in the effect of the Earth’s Magnetic field, it will experience a magnetic torque due to the magnetic lines of force passing through the area of cross-section of the loop.
The torque experienced by the loop is expressed as:
[tex]\boxed{\tau =BIA}[/tex]
Here, [tex]\tau[/tex] is the torque experienced, [tex]B[/tex] is the magnetic field, [tex]I[/tex] is the current in the loop and [tex]A[/tex] is the area of cross-section of the loop.
The strength of the Earth’s magnetic field is [tex]5\times{10^{-5}}\,{\text{T}}[/tex] .
Substitute the values in the above expression.
[tex]\begin{aligned}1.0\times{10^{-3}}&=\left({5\times{{10}^{-5}}}\right)\timesI\times\left({\pi \times{{\left({\frac{d}{2}}\right)}^2}}\right)\\I&=\frac{{1.0\times{{10}^{-3}}}}{{5\times{{10}^{-5}}\left({\pi {{\left({\frac{{0.20}}{2}}\right)}^2}}\right)}}\\&=6.366\times{10^2}\,{\text{A}}\\\end{aligned}[/tex]
The current required in the loop to experience the given torque is [tex]\boxed{6.366\times{{10}^2}\,{\text{A}}}[/tex] or [tex]\boxed{636.6\,{\text{A}}}[/tex] .
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Answer Details:
Grade: College
Subject: Physics
Chapter: Electromagnetism
Keywords:
Earth’s magnetic field, torque, maximum torque, maximum current, through the loop, experience a modest torque, T=BIA, 636 A, wire is oriented.
