A thin uniform ring of mass m and having charge Q uniformly distributed rotates around an axis perpendicular to its plane and going through its center.The angular momentum of ring is 7.5*10^-4 kg-m^2/s. The ring is in a homogeneous magnetic field of strength 0.1T and the lines of the magnetic induction are parallel with the plane of the ring. Find torque exerted. Q/m ratio = 10^-5 Ans =3.75*10^-10 I m getting my Ans as 7.5*10^-10 Share with your friends Share 4 Sakshi Kanwar answered this Dear Student Q/m=10-5Q=10-5mangular momentum=mass×areatime=7.5×10-4Torque=NBIA/2=BQA2T=0.1×Q×L22t=0.1×10-5×7.5×10-42=3.75×10-10Hope this information will clear your doubts about topic. If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible. Regards 0 View Full Answer
