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A straight, vertical wire carries a current of 2.30 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has a magnitude of B = 0.550 T and is horizontal.
A) What is the magnitude of the magnetic force on a 1.00 cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is east? Express your answer with the appropriate units.

Respuesta :

The magnitude of magnetic force of 0.01 cm of wire at mentioned current and magnetic field is 0.01265 Newtons.

The formula to be used for calculation is -

F = IBl sin theta, where F is force, I is current, B is magnetic field strength and theta is the angle between current carrying wire and magnetic field direction.

Firstly converting the length of wire into meters.

100 cm = 1 m

1 cm = 0.01 m

Keep the values in formula -

F = 2.30 × 0.550 × 0.01 × sin 90

theta is 90 as the current carrying wire and magnetic field are perpendicular to each other. Keep the value of sin 90 in formula.

F = 2.30 × 0.550 × 0.01 × 1

Performing multiplication on Right Hand Side of the equation

F = 0.01265 N

Thus, the force is 0.01265 Newtons.

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