A current carrying circular loop of radius 2 cm with unit normal n= k+ i 2 is placed in a magnetic field, B=B0 (3 i+2 k ). If B0=4 10-3 T and current I=100 2 A, the torque experienced by the loop is Wb.A. ( =3.14)
- 16 10-5 k
- 5024 10-7 k
- 5024 10-7 i
- 5024 10-7 j
Answer
(4) 5024 10-7 j
Solution
Area A= r2=3.14 (0.02)2=1.256 10-3 m2
m=IA n=100 2 1.256 10-3 i+ k 2=0.1256 ( i+ k ) A m2
= m B=0.1256 4 10-3 ( i+ k ) (3 i+2 k )
( i+ k ) (3 i+2 k )=2 ( i k )+3 ( k i )=-2 j+3 j= j
=5.024 10-4 j=5024 10-7 j Wb.A
Answer: (4)