A 5 mg particle carrying a charge of 5 10-6 C is moving with velocity of (3 i+2 k ) 10-2 m/s in a region having magnetic field B=0.1 k Wb/m2. It moves a distance of meter along k when it completes 5 revolutions. The value of is .
Answer
2
Solution
v= (3 i+2 k ) 10-2 m/s and B is along k
v =3 10-2 m/s gives circular motion; v =2 10-2 m/s is unchanged along k (helical path)
m=5 mg=5 10-6 kg
T= 2 mqB= 2 5 10-65 10-6 0.1=20 s
Distance along k in 5 revolutions =v 5T=2 10-2 100=2 m
=2
Answer: 2