Question

A 20cm long thin conducting bar travels through a uniform magnetic field at 50m/s/

A. voltage of 0.7 V is measured across the ends of this bar. What is the magnetic field magnitude here?

B. Which end of the bar has the highest electric potential? Defend your answer.

3. A 20-cm long thin conducting bar travels through a uniform magnetic field at 50 m/s. х В in) х х х х. х X х х х х х х х х х х х 30° х х х х х A. A voltage of 0.7 V is measured across the ends of this bar. What is the magnetic field magnitude here? B. Which end of this bar has the highest electric potential? Defend your answer.

Answer 1

Ans (a)

If a bar of length l is moving with velocity v in a magnetic field B ( B is perpendicular to plane of moving bar ) then induced emf is given as e =Bvl

Given

e= 0.7 V

B=?

v=50 m/s

l=20cm =0.2 m

Hence B= e/(vl)=0.7/(50×0.2)=0.007 T

(b)

If a charge q is moving with velocity v in magnetic field B then force on charge is given as F=qvB sin( theta)

Where theta is angle between B and v.

Hence using above mentioned concept positive cahrge will experience a Force in upward direction and negative charge in downward direction . Hence Upper end having higher potential then of lower one.