A coil of 160 turns and area 0.20 m² is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T.The magnetic field changes uniformly from 0.40 T in the +x direction to 0.40 T in the -x direction in 2.0 s.If the resistance of the coil is 16 Ω,at what rate is power generated in the coil? A) 5.0 W B) 10 W C) 15 W D) 20 W E) 25 W

Question

Quizplus · Accepted Answer

The magnetic field is changing uniformly, so we can use the formula for the induced EMF in a coil:
$\varepsilon = -N\frac{\Delta \Phi}{\Delta t}$
where $N$ is the number of turns, $\Delta \Phi$ is the change in magnetic flux through the coil, and $\Delta t$ is the time over which the change occurs.

The magnetic flux through the coil is given by:
$\Phi = BA\cos	heta$
where $B$ is the magnitude of the magnetic field, $A$ is the area of the coil, and $	heta$ is the angle between the magnetic field and the normal to the coil. Since the axis of the coil is parallel to the magnetic field, $	heta = 0$ and $\cos	heta = 1$.

The initial flux through the coil is:
$\Phi_i = 0.40	ext{ T}\cdot0.20	ext{ m}^2 = 0.08	ext{ Wb}$

The final flux through the coil is:
$\Phi_f = -0.40	ext{ T}\cdot0.20	ext{ m}^2 = -0.08	ext{ Wb}$

The change in flux is:
$\Delta \Phi = \Phi_f - \Phi_i = (-0.08	ext{ Wb}) - (0.08	ext{ Wb}) = -0.16	ext{ Wb}$

The time over which the change occurs is $\Delta t = 2.0	ext{ s}$.

The induced EMF in the coil is:
$\varepsilon = -N\frac{\Delta \Phi}{\Delta t} = -(160)\frac{-0.16	ext{ Wb}}{2.0	ext{ s}} = 12.8	ext{ V}$

The current in the coil is:
$I = \frac{\varepsilon}{R} = \frac{12.8	ext{ V}}{16\Omega} = 0.8	ext{ A}$

The rate at which power is generated in the coil is:
$P = I^2R = (0.8	ext{ A})^2(16\Omega) = 10.24	ext{ W}$

Therefore, the best choice is B) 10 W.

A Coil of 160 Turns and Area 0