wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A uniform magnetic field B=(20^i30^j+50^k) T exists in the space. A charge particle with specific charge 10319 C/kg enters this region at time t=0 with a velocity v=(20^i+50^j+30^k) m/s. The pitch of the helical path for the motion of the charged particle will be:

A
π100 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
π125 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
π215 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
π250 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D π250 m
The pitch of the helical path is given by

P=v||×T=(vcosθ)T

The given equation can be written as:

vcosθ=BvcosθB=Bv|B|

Now substituting vcosθ, we get;

P=Bv|B|×T

Here, T=2πmqB

P=Bv|B|×2πmqB

Here,
|B|=B=400+900+2500=3800 T

and qm=10319 C/kg

P=(20^i30^j+50^k)(20^i+50^j+30^k)3800×38π103(3800

P=(4001500+1500)(3800)2×38π1000

P=π250 m

Hence, option (d) is the right choice.

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon