A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50 T experiences a force of what magnitude

It is 0, there is no force when traveling parallel.

F = B I L

if the current is perpendicular to the B-field.

In this case the current is parallel the field, and there is zero force.

It's a trick question

A 2.0 m wire segment carrying a current of 0.60 A oriented perpendicular to a uniform magnetic field of 0.50 T experiences a force of what magnitude?

Well, well, well. Looks like we've stumbled upon a wire that couldn't resist the magnetic charm of a uniform magnetic field. How intriguing! Now, to uncover the magnitude of the force experienced by this wire segment, we can whip out our trusty formula. Are you ready? Here it comes: F = BIL, where F represents force, B represents the magnetic field, I represents the current, and L represents the length of the wire segment. Let's plug in the numbers and do some math together.

F = (0.50 T) x (0.60 A) x (2.0 m)

Now, if my calculations are correct (which they usually are, except when I divide by zero), you should find that the force experienced by this wire segment is 0.60 Newtons. Ta-da! The force has been unveiled, and the wire can finally live out its magnetic adventure. Enjoy the show!

To calculate the force experienced by a wire segment carrying a current in a magnetic field, you can use the formula for the magnetic force on a current-carrying wire. The formula can be written as:

F = BIL

Where:
F is the force experienced by the wire segment (in Newtons)
B is the magnetic field strength (in Tesla)
I is the current flowing through the wire segment (in Amperes)
L is the length of the wire segment (in meters)

In this case, you have:
B = 0.50 T (the magnetic field strength)
I = 0.60 A (the current flowing through the wire segment)
L = 2.0 m (the length of the wire segment)

Now, let's plug in the given values into the equation:

F = (0.50 T) * (0.60 A) * (2.0 m)

First, multiply the current (0.60 A) by the length (2.0 m):
F = (0.50 T) * (1.20 A·m)

Finally, multiply the magnetic field (0.50 T) by the current times the length (1.20 A·m):
F = 0.60 N

Therefore, the wire segment experiences a force of 0.60 Newtons.

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