CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A circular coil of wire consisting of 100 turns each of radius 9 cm carries of 0.4A. The magnitude of magnetic field at the centre of the coil is:


A
2.4×104 T
loader
B
3.5×104 T
loader
C
2.97×104 T
loader
D
3×104 T
loader

Solution

The correct option is B $$2.97 \times 10^{-4} \ T$$
Here $$N = 100$$

$$R = 9 cm = 9 \times 10^{-2} m $$ and $$I = 0.4 A$$

Now, $$B = \dfrac{\mu_0 NI}{2R} = \dfrac{2 \pi \times 10^{-7} \times 100 \times 0.4}{9 \times 10^{-2}}$$

$$= \dfrac{2 \times 3.14 \times 0.4}{9} \times 10^{-3} = 0.279 \times 10^{-3} T = 2.79 \times 10^{-4} T$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image