a wire having mass per unit length of 0.505 g/cm carries a 2.94 a current horizontally to the south. what are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?



Answer :

The direction of the magnetic field is east and its magnitude is 0.171T.

Here we have to find the direction and magnitude of the minimum magnetic field needed.

The given values:

mass/ length (m/l) = 0.505 g/cm

Current(I) = 2.94A

Let the magnetic field be B

Formula:

Form F = I( l× B)

As it is in the south direction. So gravity is in a downward direction.

F = I( lB)

F/l = IB

F/l = mg/l

=( 0.505× [tex]10^{-3}[/tex]/ [tex]10^{-2}[/tex]) 10

 =0.505

Now put in the equation we have:

0.505 = 2.94 × B

B = 0.171T in the east direction.

Therefore the direction is in the east and the magnitude of the magnetic field is 0.171T.

To know more about the magnetic field refer to the link given below:

https://brainly.com/question/1691855

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