Question

A straight 1.0-m long wire is carrying a current. The wire is placed perpendicular to a magnetic field of strength 0.20 t. If the wire experiences a force of 0.60 n, what is the current in the wire

Answer 1

Use the equation equation, F= q( VxB), where F is the force in Newton, V is velocity of a charged particle in m/s, q is the charge in coulombs, and B is the field in Telsa.

Replace V= distance/ time. Distance= 1.0 m

Now the equation becomes F= q/t ( DxB)

q/t= current, I

so F= I (DB sin theta) [cross product] since angle = 90 degrees, sin= 1

0.6= I (1x0.2)

solving, I= 0.6/0.2= 3A

020

I hope this will help uh

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