Question

Interactive Solution 22.25 offers some help for this problem. A copper rod is sliding on two conducting rails that make an angle of 19o with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 m/s. A 0.41-T uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle ABC during the 4.9-s period after the rod has passed point A.

Answer 1

Answer:

The magnitude of the average electromagnetic force induced in the triangle ABC after 4.9s will be 0.249V.

Explanation:

The magnitude of the emf can be obtained with the Faraday’s and Lenz’s Law:

$|emf|=|-\frac{d\phi_m}{dt}|$

In this case, the magnetic flux is:

$\phi_m=\displaystyle\int_{S}\vec{B} \,\vec{ds}=\int_{S}B_0\vec{z} \,\vec{z}ds=B_0\int_{S}\,ds=B_0S_{ABC}$

The area of the triangle ABC can be easily obtained. The base of the triangle depends linearly with the time ($b=\int_{t} v\,dt=vt$). The hight of the triangle can be obtained as a relation between sides of the triangle and the angle:

$h=btan(\alpha)=vt\cdot tan(\alpha)$

So the triangle area would be:

$S_{ABC}=\displaystyle \frac{bh}{2} =\frac{v^2t^2tan(\alpha)}{2}$

Therefore the emf results:

$|emf|=\displaystyle |\frac{d\phi_m}{dt}|=|\frac{d}{dt}(B_0\frac{v^2t^2tan(\alpha)}{2})|=B_0v^2t\cdot tan(\alpha)=0.249V$