Electro Magnetic Induction And Alternating Currents Question 477
A straight conducting metal wire is bent in the given shape and the loop is closed. Dimensions are as . Now the assembly is heated at a constant rate $ ~dT/dt=l,{}^\circ C/s $ . The assembly is kept in a uniform magnetic field B=1 T, perpendicular into the paper. Find the current in the loop at the moment, when the heating starts. Resistance of the loop is $ 10\Omega $ at any temperature. Coefficient of linear expansion $ \alpha ={{10}^{-6}}/{}^{o}C $ .
Options:
A) $ 1.5\times {{10}^{-6}},A $ anticlockwise
B) $ 1.5\times {{10}^{-6}},A $ clockwise
C) $ 0.75\times {{10}^{-6}},A $ anticlockwise
D) $ 0.75\times {{10}^{-6}},A $ clockwise
Show Answer
Answer:
Correct Answer: A
Solution:
Rate of change of area of the loop $ \frac{dA}{dt}=B \frac{dB}{dt} $ ,
$ \beta \frac{dT}{dt}=A.(2\alpha )\frac{dT}{dt}=\frac{3}{4}\times 2\times {{10}^{-6}}\times 1 $ $ =1.15\times {{10}^{-5}}m^{2}/s $
$ emf=-\frac{d\phi }{dt}=-\frac{B .dA}{dt}=-1.5\times {{10}^{-6}}V $ current in the loop $ =1.5\times {{10}^{-3}}A $ The direction will be anticlockwise as the induced current will try to negate the increase in flux due to increase in area.
BETA
