Question

One long wire carries current 32.0 A to the left along the x axis. A second long wire carries current 58.0 A to the right along the line (y = 0.280 m, z = 0).

(a) Where in the plane of the two wires is the total magnetic field equal to zero?

(b) A particle with a charge of −2.00 µC is moving with a velocity of 150î Mm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (Ignore relativistic effects.)

(c) A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.

Answer 1

(a) We know that the magnetic field due to current carrying long wire
B = _o*I /(2a)
where a is the distance from the wire
The magnetic field will be zero below the lower wire as shown in the figure , say point P
Now magnetic field due to lower wire
B₁ = (_o /2)(I₁ /x
= (_o /2)(32 /x) ----------(1)
And due to upper wire
B₂ = (_o /2)[I₂ /(x+0.28) ]
= -(_o /2)[58 /(x+0.28) ]-----------(2)
-ve sign indicate that field is toward inside the paper
Since the direction of both field will be opposite therfore
B₁ + B₂ = 0
B₁ = -B₂
(32/x)= (58/(x+0.28))
on solving
x = 0.345 m
hence at this distance magnetic field will be zero.
(b) Similarly we will calculate the magnetic field at point Q
B = B₁ + B₂
Now both wire will provide the magnetic field toward inside the paper
B = (_o /2)*[(32/0.1)+(58/0.18)]
B = 12.84*10^-5 T
Now the force ,
F = qvB  
where q is charge , v is velocity and B is the magnetic field
F = 2*10^-6*(150*10⁶)*12.84*10^-5 = 3.852*10^-2 N
(c) We know that the magentic force will be in upward direction (i.e z )
therefore electric field will be also in the z direction so charge will remain undeflected
we know that electric force is qE
F = qE
E = F/q = 3.852*10^-2 /(2*10^-6) = 1.926*10⁴ N/C