Answer to Question #108630 in Electricity and Magnetism for Abhishek

Question

a wire loop of resistance 10 ohm and radius 10cm is kept in the plane of its paper in a uniform magnetic field b.the direction of b is perpendicular to the plane of the page and points out of it and it's magnitude is increasing at the rate of 0.5t/s. determine the magnitude and direction of the induced current in the loop

Answer 1

Solution: Faraday's law states that the EMF is

(1) $\varepsilon=-\frac{d\varPhi}{dt}$ , where $\varPhi=\vec B\cdot \vec S$ is a vector dot product of magnetic field and  an element of surface area. In our case $\vec B || \vec S$ , thus $cos \hat {BS}=1$ and $\vec B\cdot \vec S=B\cdot S$

Since the area of the wire loop is constant and the frame is stationary

(2)$\frac{d\varPhi}{dt}=\frac{dB}{dt}S$ , where $S=\pi r^2=3.14\cdot (0.1m)^2=3.14\cdot 10^{-2}m^2$ . We have from (1) and (2)

(3) $\varepsilon=-\frac{dB}{dt}S=- 0.5T/s\cdot 3.14\cdot 10^{-2}m^2=-1.57\cdot 10^{-2}V$

A minus sign means that the EMC is directed opposite to the right-hand-rule [1], i.e. as shown in the figure 1.

fig 1.

fig 2.

In other words, the induction current is always directed so as to prevent changes in the magnetic flux.The induced current in the loop is

(4) $I=\frac{\varepsilon}{R}=\frac{1.57\cdot 10^{-2}V}{10 \Omega}=1.57\cdot 10^{-3}A=1.57 mA$

Answer: The magnitude of the induced current in the loop is 1.57mA and its direction is seen in figure 2.

[1] https://en.wikipedia.org/wiki/Right-hand_rule