a particle with positive charge q is a distance d from a long straight wire that carries a current i , the particle moves with speed v prependicular to the wire . what is the direction and magnitude of force when particle move towards wire
Answers
Explanation:
12th
Physics
Moving Charges and Magnetism
Problems on Force on Parallel Current Carrying Conductors
A long straight wire carrie...
PHYSICS
avatar
Asked on November 22, 2019 by
Kalpa Balla
A long straight wire carries a current i. a particle having a positive charge q and mass m, kept at distance x
0
from the wire is projected towards it with speed v. If the closest distance of approach of charged particle to wire is x
min
=x
0
e
−Yπmv/μ
0
qi
. Find Y.
168887
HARD
Help best friend
Study later
ANSWER
Magnetic field due to long wire is:
B
=−
2πx
μ
0
i
k
^
Let velocity of the charged particle at any instant be:
v
=(v
x
i
^
+v
y
j
^
)
Magnetic force on the charge,
F
=q(
v
×
B
)=−
2πx
μ
0
qiv
y
i
^
+
2πx
μ
0
q
j
^
......(i)
From Eq. (i), y component of acceleration is:
a
y
=
dt
dv
y
=
2πmx
μ
0
qiv
x
=
2πmx
μ
0
q
i
dt
dx
At minimum separation, velocity of particle is −v
j
^
(closest distance will be when velocity becomes parallel to wire or in -ve y-direction)
∫
0
−v
dv
y
=
2πm
μ
0
qi
∫
x
0
x
min
x
dx
⇒−v=
2πm
μ
0
qi
ln
∣
∣
∣
∣
∣
∣
x
0
x
min
∣
∣
∣
∣
∣
∣
Thus, x
min
=x
0
e
−2πmv/μ
0
qi
pls mark as brainlist