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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;
[tex]emf = IR = NA\frac{dB}{dt}\\\\\frac{dB}{dt} = \frac{IR}{NA} \\\\Where; A \ is \ area \ of \ the \ coil = \frac{\pi D^2}{4}\\\\[/tex]
[tex]\frac{dB}{dt}[/tex] is the rate of increase of the magnetic field, B
R is the resistance of the copper wire
[tex]R = \frac{\rho L}{A} = \frac{\rho D}{A}= \frac{\rho D}{\frac{\pi d^2}{4} }[/tex]
Where; ρ is the resistivity of copper wire = 1.68 x 10⁻⁸ Ωm
[tex]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[/tex]
[tex]\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[/tex]
Therefore, the rate of increase of the magnetic field is 0.0311 T/s.