(adsbygoogle = window.adsbygoogle || []).push({}); Torque on a soleniod / coil in magnetic field

1. The problem statement, all variables and given/known data

A solenoid having 180 turns and a cross-sectional area of 6.68 cm^{2}carries a current of 1.12 A

1) If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented perpendicular to the field.

2) If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented at 45.0* with the field

2. Relevant equations

τ = IBAsinθ

3. The attempt at a solution

1) know the following:

N=180 , B = 1.12T , I = 1.12A , θ = 90* , area = .0668 m2

τ = IBAsinθ

τ = (1.12 A) (1.12 T) ( .0668 m2) (sin90) = 0.084

I then take that and multiply by number of turns = 0.084 x 180 = 15.08

But my answer is wrong. where am I going wrong?

2) N=180 , B = 1.12T , I = 1.12A , θ = 45* , area = .0668 m2

τ = IBAsinθ

τ = (1.12 A) (1.12 T) ( .0668 m2) (sin45) = 0.060

I then take that and multiply by number of turns = 0.060 x 180 = 10.67

This is also a wrong answer. Where am I going wrong here?

Please help

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# Homework Help: Torque on a soleniod / coil

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