Moving Charges and Magnetism · Question 34

Consider a long thin conducting wire carrying a uniform current I. A particle having mass "m" and charge "q" is released at a distance "a" from the wire with a speed v0 along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [ 0 is vacuum permeability]

1. a [1 - mv02q 0 I ]
2. ae- 4 m v0q 0 I
3. a [1 - mv0q 0 I ]
4. a2

Answer

(2) ae- 4 m v0q 0 I

Solution

Take the wire along the z-axis (current along +z) and the particle in the x-z plane at distance r.

Field at the particle: B = 0 I2 r, along +y

Force F = q v B with v = vr i + vz k

Fz = qvrB = q drdt 0 I2 r

m dvz = q 0 I2 drr

Integrating from r = a (vz = v0) to r = x:

m(vz - v0) = q 0 I2 xa

Magnetic force does no work, the speed always remains v0.

At the turning point vr = 0, so |vz| = v0; the velocity has reversed, vz = -v0

-2mv0 = q 0 I2 xa

xa = - 4 m v0q 0 I

x = ae- 4 m v0q 0 I

Answer: (2)