Question

A charged particle A of charge q = 2 C has velocity v = 100 m/s. When it passes through point A and has velocity in the direction shown. The strength of magnetic field at point B due to this moving charge is (r = 2 m).the agle between them is 30.

42293_a3c6da88eb52467ead7e1a4036060562.png

A
2.5μT
B
5.0μT
C
2.0μT
D
None
Solution
Verified by Toppr

μ04πqv×rr3=μ04πqvsinθr2
=μ04π2×100×sin3004
=107×25

Was this answer helpful?
3
Similar Questions
Q1
A charge particle of charge q and mass m is moving with velocity v as shown in figure in a uniform magnetic field B along negative z-direction. Select the correct alternative(s).

View Solution
Q2

A charged particle (charge q, mass m) has velocity v0 at the origin in +x direction. In space, there is a uniform magnetic field B in –z direction. Find the y coordinate of the particle when it crosses y axis.

View Solution
Q3

A charged particle of charge q and mass m is moving with velocity v (as shown in the figure) in a uniform magnetic field B along negative z – direction. Select the correct alternative(s).

View Solution
Q4

A charged particle (charge q, mass m) has velocity v0 at the origin in +x direction. In space, there is a uniform magnetic field B in –z direction. Find the y coordinate of the particle when it crosses y axis.

View Solution
Q5

Six point charges, each of the same magnitude q, are arranged in different manners as shown in List 2 .In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest.Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B the magnetic field at M and μ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.

View Solution
Solve
Guides