A wire of length 0.25 m is conducting a current of I = 7.9 A in the x direction

Question

A wire of length 0.25 m is conducting a current of I = 7.9 A in the x direction through a B = 4.0-T uniform magnetic field that is directed parallel to the wire in the -x direction. What are the magnitude and direction of the magnetic force on the wire?

The magnetic field changes, so that now the force on the wire is F = 7.0 N in the y direction. What are the magnitude and direction of the magnetic field now? (Assume that the magnetic field B is perpendicular to I.)

A wire of length 0.25 m is conducting a current of I 7.9 A in the +x direction through a B-40-T uniform magnetic field that is directed parallel to the wire in the -xdirection. What are the magnitude and direction of the magnetic force on the wire? Number +X -X O none of the above + x

Explanation / Answer

Solution:

a) Using F = ILB sin

where is the angle between current element IL and B (magnetic field)

Now, According to question magnetic field and current both are antiparallel So , = 180

Using the formula F = ILBsin180 (sin 180 = 0)

=> F = 0 N

So, F = 0 N

Direction = None of the above

b) Again using F = ILBsin

=> F = I L B (taking sin =1)

=> 7 = 7.9 * 0.25 * B

=> B = 3.544 T

Direction = +Z direction ( Using Right hand rule)

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