Finding Total B-Field and Direction in a Wire-Magnet System

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The discussion focuses on calculating the total magnetic field at a specific point in a wire-magnet system. A straight wire carrying a current of 8.00 A in the -y-direction generates a magnetic field, while a uniform magnetic field of 1.50 x 10^-6 T exists in the +x-direction. The magnitude of the magnetic field from the wire is correctly calculated as 1.6 x 10^-6 T. Participants emphasize that the total magnetic field must be determined by vector addition, considering both the horizontal and vertical components. To find the direction of the total field, trigonometric functions are suggested to calculate the angle formed by the resultant vector.
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Homework Statement



A long, straight wire lies along the y-axis and carries a current I = 8.00 A in the -y-direction . In addition to the magnetic field due to the current in the wire, a uniform magnetic field with magnitude is in the +x-direction.

What is the magnitude of the total field at the point x = 0, z = 1.00 m in the xz-plane?

What is its direction? from x to z axis.

Homework Equations



Using magnetic field equation B = (mu_0 *I)/2*pi*r

The Attempt at a Solution



B = (mu_o*I)/2*pi*r = (4pi x 10^-7)(8.00A)/2pi*(1.00m) = 1.6 x 10^-6 T

using the same equation B = (mu_o*I)sin(theta)/2*pi*r, but I am not sure how to continue this problem.

Can someone help me how to calculate for the direction for this B-field that goes from x to z axis?

Thanks
 

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I can't see the attachment yet, but I'll try to offer as much advice as I can without it.

You have found the magnitude of the B field from the wire correctly. You also didn't give the magnitude of the constant field in the x direction. I assume you forgot to put it in, since its seems to be something that needs to be given. Here are some hints to help you get to the answer.

Hints:

What direction does the B field from the wire point in?

If you know this then you know the two B vectors, both magnitude and direction. Now the total B field will be their sum, correct? How do you add these two vectors?
 
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G01 said:
I can't see the attachment yet, but I'll try to offer as much advice as I can without it.

You have found the magnitude of the B field from the wire correctly. You also didn't give the magnitude of the constant field in the x direction. I assume you forgot to put it in, since its seems to be something that needs to be given. Here are some hints to help you get to the answer.

Hints:

What direction does the B field from the wire point in?

If you know this then you know the two B vectors, both magnitude and direction. Now the total B field will be their sum, correct? How do you add these two vectors?

Thank for the advice.

From the RHR, the magnetic field is pointing in the x-axis direction, similar to B_o.

The value for B_o is given as 1.50 x 10^-6 T, so the total B field will then be 1.50 x 10^-6 + 1.6 x 10^-6 T = 1.0 x 10^-7 T.

I am still confuse on how to find the angle, can you further explain?

Thanks
 
whitetiger said:
The value for B_o is given as 1.50 x 10^-6 T, so the total B field will then be 1.50 x 10^-6 + 1.6 x 10^-6 T = 1.0 x 10^-7 T.

This is not the correct sum of the vectors. Remember Vectors don't add like real numbers. Here you have a horizontal component, and a vertical component. The sum vector will be the hypotenuse of the triangle they form. Does this help you remember how to find the sum? Then, use trig to find the angle. Does this help?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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