First slide

Magnetic flux and induced emf

Question

Plane figures made of thin wires of resistance R = 50 milli ohm/meter are located in a uniform magnetic field perpendicular to the plane of the figures and which
decrease at the rate dB/dt = 0.1 mT/s. Then currents in the inner and outer boundary are (The inner radius a = 10 cm and outer radius b = 20 cm)

Moderate

A

$10^{- 4} A (Clockwise), 2 \times 10^{- 4} A (Clockwise)$
B

$10^{- 4} A (Anticlockwise), 2 \times 10^{- 4} A (Clockwise)$
C

$2 \times 10^{- 4} A (clockwise), 10^{- 4} . A (Anticlockwise)$
D

$2 \times 10^{- 4} A (Anticlockwise), 10^{- 4} A (Anticlockwise)$

Solution

$Current in the inner coil i_{1} = \frac{ε}{R} = \frac{A_{1}}{R_{1}} \frac{d B}{d t}$
Length of the inner coil $= 2 π a$
So its resistance $R_{1} = 50 \times 10^{- 3} \times 2 π (a)$
$\begin{array}{l} ∴ i_{1} = \frac{π a^{2}}{50 \times 10^{- 3} \times 2 π (a)} \times 0.1 \times 10^{- 3} = 10^{- 4} A \\ According to Lenz's law, direction of i_{l} is clockwise. \\ Induced current in outer coil i_{2} = \frac{ε_{2}}{R_{2}} = \frac{A_{2}}{R_{2}} \frac{d B}{d t} \\ \Rightarrow i_{2} = \frac{π b^{2}}{50 \times 10^{- 3} \times (2 π b)} \times 0.1 \times 10^{- 3} = 2 \times 10^{- 4} A (CW) \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App