Question

The magnitude of the earth's magnetic field at a place$B_0$ and the angle of dip is$\delta$. A horizontal conductor of length$l$ lying along the magnetic North-South moves Eastwards with a velocity v, the emf induced across the conductor is

• A. zero
• B. $B_{0}lv\sin \delta$
• C. $B_{0}lv$
• D. $B_{0}lv\cos \delta$
• E. $B_0\sin \delta$

Answer 1

The correct option is B

$B_{0}lv\sin \delta$

Step 1. Given data

The magnitude of the earth's magnetic field at a place$B_0$.

The angle of dip$=\delta$.

Step 2. Formula used

Motional-induced EMF is given by,

$e=Bvl$, where($B=$Magnetic field intensity, $v=$Velocity of the conductor, $l=$Length of the conductor)

Step 3. Calculating the induced EMF

The horizontal conductor of length $l$ lies along the north-south direction when it moves in the east direction with a velocity$v$ that is horizontally normal to $B_v$

The EMF will get induced and the magnetic field in the east direction starts acting according to Fleming's right-hand rule, the direction of induced emf is along $B_v$.

Thus the induced emf will be,

$\begin{array}{rcl}E& =& B_{vertical}vl\\ B_v& =& B_0\sin \delta \\ E& =& B_{0}vl\sin \delta \end{array}$

Thus, the EMF across the conductor is $B_{0}lv\sin \delta$.

Hence, option B is the correct answer.