A circular coil of 20 turns and radius 5.0 cm is placed with its plane oriented at 90° to a uniform $\vec { B }$ field of $0.10 \mathrm {~T}$ The field is now increased at a steady rate,reaching a value of $0.50 \mathrm {~T}$ after 4.0 seconds.What emf is induced in the coil?
A)0.016 V
B)0.021 V
C)0.026 V
D)0.031 V
E)0.036 V

Question

Quizplus · Accepted Answer

The magnetic flux through the coil is given by:$$\Phi = BA\cos\theta$$where:* $B$ is the magnetic field strength* $A$ is the area of the coil* $\theta$ is the angle between the magnetic field and the normal to the coilIn this case, $B = 0.10 \mathrm {~T}$, $A = \pi r^2 = \pi (0.05 \mathrm {~m})^2 = 7.85 \times 10^{-3} \mathrm {~m}^2$, and $\theta = 90^\circ$. Therefore, the initial magnetic flux is:$$\Phi_i = BA\cos\theta = (0.10 \mathrm {~T})(7.85 \times 10^{-3} \mathrm {~m}^2)(\cos 90^\circ) = 0 \mathrm {~Wb}$$After 4.0 seconds, the magnetic field strength has increased to $B = 0.50 \mathrm {~T}$. Therefore, the final magnetic flux is:$$\Phi_f = BA\cos\theta = (0.50 \mathrm {~T})(7.85 \times 10^{-3} \mathrm {~m}^2)(\cos 90^\circ) = 3.93 \times 10^{-3} \mathrm {~Wb}$$The emf induced in the coil is given by:$$\varepsilon = -\frac{d\Phi}{dt}$$where:* $\varepsilon$ is the emf* $\Phi$ is the magnetic flux* $t$ is the timeIn this case, the magnetic flux is changing at a constant rate of:$$\frac{d\Phi}{dt} = \frac{\Phi_f - \Phi_i}{t} = \frac{3.93 \times 10^{-3} \mathrm {~Wb} - 0 \mathrm {~Wb}}{4.0 \mathrm {~s}} = 9.83 \times 10^{-4} \mathrm {~Wb/s}$$Therefore, the emf induced in the coil is:$$\varepsilon = -\frac{d\Phi}{dt} = -(9.83 \times 10^{-4} \mathrm {~Wb/s}) = \boxed{0.016 \mathrm {~V}}$$

A Circular Coil of 20 Turns and Radius 5 B⃗\vec { B }B

A Circular Coil of 20 Turns and Radius 5 $\vec { B }$