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?

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1 year 2021-08-15T01:26:55+00:00 1 Answers 245 views 0

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.