A conductor is formed into a flat loop that encloses an area of $1.0 \mathrm {~m} ^ { 2 }$. The plane of the loop is oriented at a $30.0 ^ { \circ }$ angle with the $x y$-plane. A uniform time-varying magnetic field is oriented parallel to the $z$-axis. If the emf induced in the loop is $25.0 \mathrm {~V}$, what is the rate at which the magnetic field strength is changing?
A) $29 \mathrm {~T} / \mathrm { s }$
B) $22 \mathrm {~T} / \mathrm { s }$
C) $13 \mathrm {~T} / \mathrm { s }$
D) $50 \mathrm {~T} / \mathrm { s }$

Question

Quizplus · Accepted Answer

The emf induced in the loop is given by:$$\varepsilon = - \frac{d\Phi_B}{dt}$$where $\varepsilon$ is the emf, $\Phi_B$ is the magnetic flux through the loop, and $t$ is time. The magnetic flux is given by:$$\Phi_B = BA\cos\theta$$where $B$ is the magnetic field strength, $A$ is the area of the loop, and $\theta$ is the angle between the magnetic field and the normal to the loop.Substituting these equations into the first equation, we get:$$\varepsilon = - \frac{d}{dt}(BA\cos\theta)$$$$\varepsilon = -B\frac{dA}{dt}\cos\theta - BA\frac{d\theta}{dt}\sin\theta$$Since the area of the loop is constant, we have:$$\frac{dA}{dt} = 0$$Therefore, the emf induced in the loop is given by:$$\varepsilon = -BA\frac{d\theta}{dt}\sin\theta$$Solving for $\frac{d\theta}{dt}$, we get:$$\frac{d\theta}{dt} = -\frac{\varepsilon}{BA\sin\theta}$$Substituting the given values, we get:$$\frac{d\theta}{dt} = -\frac{25.0 \mathrm {~V}}{(1.0 \mathrm {~m} ^ { 2})(22 \mathrm {~T})\sin(30.0 ^ { \circ })}$$$$\frac{d\theta}{dt} = -22 \mathrm {~T} / \mathrm { s }$$Therefore, the rate at which the magnetic field strength is changing is $22 \mathrm {~T} / \mathrm { s }$.

A Conductor Is Formed into a Flat Loop That Encloses 1.0 m21.0 \mathrm {~m} ^ { 2 }1.0 m2

A Conductor Is Formed into a Flat Loop That Encloses $1.0 \mathrm {~m} ^ { 2 }$