Question

A coil of 200 turns has a cross-section area 900mm². It carries a current of 2A. The plane of coil is perpendicular to a uniform m.f. of 0.5T. Calculate: - (a) magnetic moment of coil. (b) The torque acting on the coil.

Answer 1

$\rule{200}{2}$

$\huge\boxed{\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R:-}}}}}$

Given: mean radius = 15 cm = 0.15 m, Number of turns = 3000, the relative permeability = μr = 1000, current through coil = i = 2 A, μo = 4π x 10-7 Wb/Am.

To Find: Magnetic field = B =?

Solution:

n = N/2πr = 3000 / (2π x 0.15)

∴ B = μr μ0 n i = 1000 x 4π x 10-7 x (3000 / (2π x 0.15) x 2

∴ B = 1000 x 2 x 10-7 x (3000 /0.15) x 2 = 1000 x 2 x 10-7 x (20000) x 2

∴ B = 8 T

Ans: Magnetic field = 8 T

Answer 2

Answer:

Given: mean radius = 15 cm = 0.15 m, Number of turns = 3000, the relative permeability = μr = 1000, current through coil = i = 2 A, μo = 4π x 10-7 Wb/Am.

To Find: Magnetic field = B =?

Solution:

n = N/2πr = 3000 / (2π x 0.15)

∴ B = μr μ0 n i = 1000 x 4π x 10-7 x (3000 / (2π x 0.15) x 2

∴ B = 1000 x 2 x 10-7 x (3000 /0.15) x 2 = 1000 x 2 x 10-7 x (20000) x 2

∴ B = 8 T

Ans: Magnetic field = 8 T

Explanation:

hope it will be help you