A circular coil of diameter 10 cm has 10 turns and carries a current of 5A. It is placed in a uniform field of 0.5 T with its plane parallel to the field. The torque on the coil in Nm, is
A
(6.25π)×10−2 N-m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(6.25π)×10−3 N-m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(6.25π)×10−4 N-m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A(6.25π)×10−2 N-m
θ=90;sin90=1
T =Bi An =0.5×5×πr2×10=25×π(5×10−2)2=625π×10−4=6.25π×10−2Nm.