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a straight wire of length 1.0 m and mass 231 g carries current I in a region where the magnitude field is directed horizontal from the left to right. The wire lies perpendicular to the direction of the magnetic field and the current direction is out of the paper. If the magnitude of the magnetic field is 0.59T, what is the minimum value of the magnitude of the current I in this wire so that its weight is completely supported by magnitude force due to the magnetic field it is in.

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To determine the minimum value of the current, let's consider the forces acting on the wire. In this case, we have the gravitational force pulling the wire downwards and the magnetic force exerted on the wire due to the magnetic field.

The gravitational force is given by the equation:

F_gravity = m * g

where m is the mass of the wire and g is the acceleration due to gravity (approximately 9.8 m/s^2).

The magnetic force is given by the equation:

F_magnetic = I * L * B

where I is the current in the wire, L is the length of the wire, and B is the magnitude of the magnetic field.

In order for the weight of the wire to be completely supported by the magnetic force, the magnetic force must be equal to the gravitational force. Therefore, we can set the two equations equal to each other:

I * L * B = m * g

Now, let's plug in the given values:

I * 1.0 m * 0.59 T = 0.231 kg * 9.8 m/s^2

Simplifying the equation, we have:

I = (0.231 kg * 9.8 m/s^2) / (1.0 m * 0.59 T)

I ≈ 3.84 A

So, the minimum value of the current in the wire would be approximately 3.84 Amperes in order for its weight to be completely supported by the magnetic force due to the magnetic field it is in.
User Agost Biro
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