Test Index

NCERT Class XII Chapter
Electromagnetic Induction
Questions With Solutions

© examsnet.com
Question : 4 of 17
Marks: +1, -0
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the e.m.f developed across the cut if velocity of loop is 1 cm s−1s^{-1} in a direction normal to the (a) longer side (b) shorter side of the loop? For how long does the induced voltage last in each case?
Solution:  
Here A = 8 × 2 = 16 cm2\mathrm{cm}^2 = 16 × 10−4 m210^{-4}\,\mathrm{m}^2
B = 0.3 T
v = 1 cm s−1s^{-1} = 10−210^{-2} m s−1s^{-1}
Induced emf, ε = ?
(i) When velocity is normal to longer side,
l = 8 cm = 8 × 10−210^{-2} m
ε = Blv = 0.3 × 8 × 10−210^{-2} × 10−210^{-2} = 2.4 × 10−410^{-4} V
Time , t = distance movedvelocity\frac{\text{distance moved}}{\text{velocity}} = 2×10−210−2\frac{2\times 10^{-2}}{10^{-2}} = 2 sec
(ii) When velocity is normal to shorter side,
l = 2 cm = 2 × 10−210^{-2} m
ε = Blv = 0.3 × 2 × 10−210^{-2} × 10−210^{-2} = 0.6 × 10−410^{-4} V
Time , t = distance movedvelocity\frac{\text{distance moved}}{\text{velocity}} = 8×10−210−2\frac{8\times 10^{-2}}{10^{-2}} = 8 sec
© examsnet.com
Go to Question:
Prev Question
Next Question