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

A charge particle of charge q = 2 C has velocity v=100ms1. When it passes point A adn has velocity in the direction shown. The strength of magnetic field at the point B due to this moving charge is (r = 2m)
1280769_24689dcbae234ff8b787bf63185eaebe.png

A
2.5μT
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
5.0μT
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2.0μT
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2.5μT
Given :charge(q)=2C,v=100m/s, r=2m,θ=30v=100ms−1
Solution: We know the formula of magnetic field on moving charge is :
B=μ0M4πr3
And as we know the know the formula of magnetic moment in terms of q &v
M=q(v×r)
B=μ0q(vrsin300)4πr3=107×2×100×14×2=2.5μT
Hence the correct answer is A

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon