A 100-turn, 7.0-cm-diameter coil is made of 0.50-mm-diameter copper wire. A magnetic field is perpendicular to the coil. At what rate must B

Question

A 100-turn, 7.0-cm-diameter coil is made of 0.50-mm-diameter copper wire. A magnetic field is perpendicular to the coil. At what rate must B increase to induce a 2.0 A current in the coil?

in progress 0
niczorrrr 5 months 2021-08-15T01:26:55+00:00 1 Answers 69 views 0

Answers ( )

    0
    2021-08-15T01:28:25+00:00

    Answer:

    The rate of increase of the magnetic field is 0.0311 T/s.

    Explanation:

    Given;

    number of turns of the coil, N = 100 turn

    diameter of the coil, D = 7.0 cm = 0.07 m

    diameter of the copper wire, d = 0.5 mm = 0.5 x 10⁻³ m

    induced current, I = 2.0 A

    The induced emf is calculated as;

    emf = IR = NA\frac{dB}{dt}\\\\\frac{dB}{dt} = \frac{IR}{NA} \\\\Where; A \ is \ area \ of \ the \ coil = \frac{\pi D^2}{4}\\\\

    \frac{dB}{dt} is the rate of increase of the magnetic field, B

    R is the resistance of the copper wire

    R = \frac{\rho L}{A} = \frac{\rho D}{A}= \frac{\rho D}{\frac{\pi d^2}{4} }

    Where; ρ is the resistivity of copper wire = 1.68 x 10⁻⁸ Ωm

    R = \frac{(1.68 \times 10^{-8})(0.07)}{\frac{\pi (0.5\times 10^{-3})^2}{4}  } \\\\R = 5.989 \times 10^{-3}  \ ohms

    \frac{dB}{dt} = \frac{(2)(5.989 \times 10^{-3})}{(100)(\frac{\pi \times (0.07)^2}{4} )} \\\\\frac{dB}{dt} =0.0311 \ T/s

    Therefore, the rate of increase of the magnetic field is 0.0311 T/s.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )