Question

The horizontal component of the earth's magnetic field at a place is $3\times {10}^{-4}T$ and the dip is $\theta$ = $ta{n}^{-1}$ (4/3). A metal rod of length 0. 25 m placed in the north-south position is moved at a constant speed of 10 cm/s towards the east. The e.m.f induced in the rod will be:

• A. zero
• B. 1 mV
• C. 5 mV
• D. 10 mV

Answer 1

The correct option is B 1 mV

$\begin{array}{c}Given,\\ B_h=(3\times {10}^{-4})T\\ V=10cm/s=10\times {10}^{-2}m/s\\ \theta ={\tan }^{-1}\left(\frac{4}{3}\right)\Rightarrow \tan \theta =\frac{4}{3}\\ \therefore \theta ={53}^0\\ and,Rod\left(l\right)=\left(0.25\right)m\\ And,\\ emf=vBl\\ emf=\left(B_V\right)\left(v\right)\left(l\right)----\left(i\right)[B=B_V\\ Now,\\ \tan \theta =\frac{B_V}{B_h}\Rightarrow B_V=B_h\tan \theta \\ =(3\times {10}^{-4})\frac{4}{3}\\ =4\times {10}^{-4}T\\ Hence,from\left(i\right)\\ emf=(4\times {10}^{-4})(10\times {10}^{-2})\left(0.25\right)\\ \therefore emf=1.0\times {10}^{-3}=1mv\\ sothecorrectoptionisB.\end{array}$