question_answer

A rectangular 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 voltage developed across the cut, if velocity of loop is \[1\text{ }cm{{s}^{-1}}\] in a direction normal to the

(i) longer side

(ii) shorter side of the loop For how long does the induced voltage last in each case?

 

Answer:

            Here, area, \[A=8\times 2=16c{{m}^{2}}=16\times {{10}^{-4}}{{m}^{2}}\] \[B=0.3.T,\] \[v=1cm/s={{10}^{-2}}m/s\] Voltage developed, e = ? (i) When velocity is normal to longer side,             \[l=8cm=8\times {{10}^{-2}}m\] \[e=Blv=0.3\times 8\times {{10}^{-2}}\times {{10}^{-2}}=2.4\times {{10}^{-4}}V\] Induced emf lasts till the loop comes out of the magnetic field. Time,    \[t=\frac{\text{Distance}\,\,\text{moved}}{\text{Velocity}}\]             \[t=\frac{2\times {{10}^{-2}}}{{{10}^{-2}}}\] \[\Rightarrow \]   t = 2s (ii) When velocity is normal to shorter side,             \[l=2cm=2\times {{10}^{-2}}m\] \[\therefore \] \[e=Blv=0.3\times 2\times {{10}^{-2}}\times {{10}^{-2}}=0.6\times {{10}^{-4}}V\] The induced voltage lasts for time, \[t=\frac{\text{Distance}\,\,\text{covered}}{\text{Velocity}}=\frac{8\times {{10}^{-2}}}{{{10}^{-2}}}=8\,\,s\]