Calculating Torque & Angular Acceleration of a Current Loop

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SUMMARY

The discussion focuses on calculating the torque and angular acceleration of a current loop with an area of 1.5 cm² carrying a current of 3.6 mA in a magnetic field of 4.2 x 10^-5 T. The magnetic moment (m) is calculated as 5.4 x 10^-5 A·m² using the formula m = IA. The torque (T) is determined using T = m x B, resulting in T = 2.268 x 10^-9 N·m when substituting the Earth's magnetic field value. The angular acceleration is derived from the torque and the rotational inertia of 6.2 g·cm².

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Homework Statement


A small current loop of area A = 1.5cm^2 carries current I = 3.6mA. What torque acts on it in a lab where its magnetic moment is perpendicualr to the Earth's magnetic field of
4.2 x 10^-5
(b) Find its angular acceleration if its rotational inertia about a diameter is 6.2gcm^2


Homework Equations



T = m x B
m = IA

The Attempt at a Solution


I know that m = (.015)(.0036) = 5.4 * 10^-5.

T = abs(5.4 * 10^-5) * abs(B) * Sin(PI/2)

I'm a bit confused about what B is, should I use the B for Earth's magnetic field?
 
Physics news on Phys.org
Yes, use the given B for the Earth's magnetic field.
 

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