a question regarding torque in a magnetism setting?
I have a question regarding torque in a magnetism context. I understand that the equation τ = IABsin(θ) will be relevant to this problem, but I'm unsure how to apply it effectively. Could someone please guide me through the process?
Specifically, I need to determine the magnitude of the torque on the circular current loop shown in the figure. Additionally, what is the loop's equilibrium position? The options provided are:
- When the loop is rotated by 90 degrees around the vertical axis.
- When the loop is rotated by 90 degrees around the horizontal axis directed into the page.
- When the loop is rotated by 90 degrees around the horizontal axis directed to the right.
Thank you for your assistance!
2 Answers
Magnetic moment : m = I / A = 0.2/[π(2x10-³)²] = 1.59x10^-4 A/m²
Amperes’ law for left current-carrying conductor yields formula
for magnetic flux density:
B = μI / 2πr = (4πx10^-7)2 / 2π(0.02) = 10^-5 TTorque: T = mxB = mBsinφ = mB = (1.59x10^-4)(10^-5)
= 1.59x10^-9 N mBy the Lorentz force law we know that the force
must be perpendicular to both current direction
and magnetic field direction.
Using right hand rule: the force cause 90 degree
rotation w.r.t horizontal axis.
First find the magnetic dipole moment of the loop<mu>:
<mu>= IA = I(pi)r^2 = 0.2(pi)(0.002/2)^2 = 6.28 x 10^-7
Then find the magnetic field strength (B):
B=(mu_0)I/(2(pi)d = (4(pi) x 10 ^ -7 2 / (2(pi)*0.02) = 2 x 10 ^ -5
We know that the wire and the loop are perpendicular to each other making 90 degrees
torque = <mu> x B = (mu)B sin(angle) = (6.28 x 10^-7)(2 x 10 ^ -5) sin(90) = 1.2566 x 10 ^ -11
Be careful. Mastering Physics told me my answer was wrong with this; They wanted 1.3 x 10 ^ -11 Nm
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