First slide

Magnetic flux and induced emf

Question

A square non-conducting loop, 20 cm, on a side is placed in a magnetic field. The center of side AB coincides with the center of magnetic field. The magnetic field is increasing at the rate of 2 T/s. The potential difference between C and D is

Difficult

A

80 mV
B

zero
C

40 mV
D

60 mV

Solution

Perpendicular distance between CD and O is d = 20cm.

Consider a small element at distance x of length dx.
Induced electric field is tangential to the circle of radius $r = \sqrt{d^{2} + {(\frac{l}{2} - x)}^{2}}$ .
$\int \vec{E} . \vec{d l} = A \frac{d B}{d t} \Rightarrow E (2 π r) = (π r^{2}) \frac{d B}{d t} \Rightarrow E = \frac{r}{2} \frac{d B}{d t} N o w, △ V = \int (E \cos θ) d x \Rightarrow △ V = \int \frac{r}{2} \frac{d B}{d t} \times \frac{d}{r} \times d x \Rightarrow △ V = \frac{d}{2} \frac{d B}{d t} l_{C D} = \frac{0.2}{2} \times 2 \times 0.2 = 0.04 V = 40 m V$

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