A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10³ T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively: (Take µ0 = 4π×10−7 T m/A)
2 A, 4 A m²
2.5 A, 20 A m²
2.5 A, 2 A m²
2 A, 10 A m²
2.5 A, 2 A m²
Step 1: Understanding the Concept:
A current-carrying coil creates a magnetic field at its center and also acts as a magnetic dipole with a specific magnetic moment.
Step 2: Key Formula or Approach:
1. Magnetic field at centre (B) = µ0NI/2r 2. Magnetic moment (M) = NIA, where A=πr2
Step 3: Detailed Explanation:
Given: N =100, r =0.05 m, B =3.14×10−3 T (which is π×10−3 T). 1. Find Current (I):
3.14×10−3 = 4π× 10−7× 100× I/2×0.05
π×10−3 = 4π× 10−5× I/0.1
10−3 =4×10−4×I
I = 10−3/4×10−4 = 10/4 =2.5 A
2. Find Magnetic Moment (M):
A=πr2 =π×(0.05)2 =3.14×0.0025 m2
M=100×2.5×(3.14×0.0025)
M=250×0.00785≈1.96≈2 A m2
Step 4: Final Answer:
The current is 2.5 A and the magnetic moment is 2 A m².