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A 2. 0-m wire carrying a current of 0. 60 a is oriented parallel to a uniform magnetic field of 0. 50 t. what is the magnitude of the force it experiences?

Sagot :

The magnitude of the force is(F)= 0 N.

How can we calculate the value of the magnitude of the force?

To calculate the force here we use the formula,

[tex]F= I \times L \times B \times sin\theta[/tex]

Here we are given,

I= The current passing through the wire = 0.60A.

L= The length of the wire= 2.0 m.

B= The magnetic field created by the wire= 0.50T.

[tex]\theta[/tex]= The angle that the wire makes after orientation= 0° (because its orientation is parallel).

We have to find the magnitude of the force = F

Now, we substitute the known values in the above equation,

[tex]F= I \times L \times B \times sin\theta[/tex]

Or, [tex]F= 0.60 \times 2.0 \times 0.50 \times sin (0)[/tex]

Or, F=0

Now from the above calculation we can conclude that, the magnitude of force is 0 N.

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