Question
Question: A coil of 50 turns is situated in a magnetic field b = 0.25weber/m<sup>2</sup> as shown in figure. A...
A coil of 50 turns is situated in a magnetic field b = 0.25weber/m2 as shown in figure. A current of 2A is flowing in the coil. Torque acting on the coil will be
A
0.15 N
B
0.3 N
C
0.45 N
D
0.6 N
Answer
0.3 N
Explanation
Solution
Since plane of the coil is parallel to magnetic field. So θ = 90o
Hence τ = NBiA sin 90o = NBiA
= 50 × 0.25 × 2 × (12 × 10–2 × 10 × 10–2) = 0.3 N.