# NEET Questions Solved

The magnitude of the earth’s magnetic field at a place is B0 and the angle of dip is δ. 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

(1) Zero

(2) B0lv sin δ

(3) B0lv

(4) B0lv cos δ

(2) When a conductor lying along the magnetic north-south, moves eastwards it will cut vertical component of B0. So induced emf

$e=v{B}_{V}l=v\left({B}_{0}\mathrm{sin}\delta \text{\hspace{0.17em}}l\right)={B}_{0}l\text{\hspace{0.17em}}v\mathrm{sin}\delta$

Difficulty Level:

• 10%
• 67%
• 8%
• 18%
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