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Magnetic Field Q?
Consider a conducting ring of mass 2kg & radius 0.5m. It is placed vertical on a smooth horizontal surface. A magnetic field of flux density 10T is along the plane. The angular acceleration of the ring at the instance at which a current of 4A begins to flow, (the moment of inertia of the ring about its axis is 1/2mr^2) is?
1 Answer
- 7 years ago
torque=B*I*A
where B=magnetic field=10T
I=current=4A
A=Area of ring=pie*radius(r)^2=3.14*(0.5^2)
torque=31.4N m
and since torque=miment of inertia*angular acceleration
substituting known values
angular acceleration=31.4/(moment of inertia)
=31.4/0.5=62.8
