Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
An air cored toroidal coil has 3000 turns and carries current 0.1 A.
the length of the magnetic circuit is 30 cm and the cross-sectional area of the coil
is 4cm^2. Determine the:
(i) magnetic field strength;
(ii) magnetic flux density. (6 marks)craft1 electrical June/July 2020
i. To find the magnetic field strength, we can use the formula B = u0 * n * I / l, where B is the magnetic field strength, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, and l is the length of the magnetic circuit. Plugging in the values given, we get B = (4pi10^-7 Tm/A) * 3000 turns * 0.1 A / 30 cm = 4.210^-4 T.
ii. To find the magnetic flux density, we can use the formula B = u0 * n * I / (A * l), where B is the magnetic flux density, u0 is the permeability of free space, n is the number of turns in the coil, I is the current in the coil, A is the cross-sectional area of the coil, and l is the length of the magnetic circuit. Plugging in the values given, we get B = (4pi10^-7 Tm/A) * 3000 turns * 0.1 A / (4 cm^2 * 30 cm) = 3.110^-3 T.