How Does Magnetic Field Orientation Affect Torque on a Solenoid?

[jcvince17]

Torque on a soleniod / coil in magnetic field

Homework Statement

A solenoid having 180 turns and a cross-sectional area of 6.68 cm² 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

Homework Equations

τ = IBAsinθ

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

 

[alphysicist]

Hi jvince17,

Homework Statement

A solenoid having 180 turns and a cross-sectional area of 6.68 cm² 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

Homework Equations

τ = IBAsinθ

The Attempt at a Solution

1) know the following:
N=180 , B = 1.12T , I = 1.12A , θ = 90* , area = .0668 m2

This area is not correct. Remember that the area given in the problem was in cm², not cm. Do you see what it need to be?

 

[jcvince17]

Hi jvince17,



This area is not correct. Remember that the area given in the problem was in cm², not cm. Do you see what it need to be?

I sure do! thank you very much. I changed my area and recalculated and both answers are now correct!

 

[florence.noy]

Good evening,

This is my first time doing this since i had just completed my registration. I am really very eager to know exactly my answer to the problem because when i tried solving it for number 1 my answer is (0) because i am using (cos 90) which is equal to 0. In number 2 my answer is 0.1066522483 N.m. I do not know if I am correct. If I am wrong please I need to know where did I get it wrong.

Thank you very much and more power to us!

 

[asteriatic]

!

I would like to provide a response to the above content on torque on a solenoid/coil in a magnetic field.

Firstly, it is important to note that the torque on a solenoid or coil in a magnetic field is a result of the interaction between the magnetic field and the current flowing through the coil. The equation for torque on a solenoid/coil is τ = IBAsinθ, where I is the current, B is the magnetic field, A is the cross-sectional area, and θ is the angle between the magnetic field and the axis of the solenoid/coil.

In the given problem, there are a few errors in the calculations for both parts 1 and 2. Firstly, the given area of 6.68 cm2 should be converted to m2, which gives 0.000668 m2. Secondly, the angle θ in part 2 is given as 45.0*, which should be converted to radians (θ = π/4).

After correcting these errors, the correct calculations for part 1 would be:

τ = (1.12 A) (1.12 T) (0.000668 m2) (sin90*) = 0.000844 Nm
Multiplying this by the number of turns (180) gives a final torque of 0.152 Nm.

Similarly, for part 2:

τ = (1.12 A) (1.12 T) (0.000668 m2) (sinπ/4) = 0.000599 Nm
Multiplying this by the number of turns (180) gives a final torque of 0.108 Nm.

I hope this helps to clarify the correct approach to solving this problem. It is important to pay attention to the units and to convert them appropriately, as well as to use the correct values for the angle θ. Keep up the good work!