Given that,
Speed = 3.5 m/s
Magnetic field = 50μT
Angle = 60°
(A). We need to find the direction of magnetic force
Using formula of magnetic force
[tex]\vec{F}=q(\vec{v}\times\vec{B})[/tex]
Here, [tex](\vec{v}\times\vec{B})[/tex]= down
But , charge is negative.
So, the direction of magnetic force will be up.
(B). We need to calculate the magnetic electric field
Using formula of magnetic force
[tex]F=qvB\sin\theta[/tex]
[tex]qE=qvB\sin\theta[/tex]
[tex]E=vB\sin\theta[/tex]
Where, v = speed
B = magnetic field
Put the value into the formula
[tex]E=3.5\times50\times10^{-6}\sin60[/tex]
[tex]E=0.000151\ N/C[/tex]
[tex]E=1.5\times10^{-4}\ N/C[/tex]
Hence, (A). The direction of magnetic force is UP
(c) is correct option
(B). The magnetic electric field is [tex]1.5\times10^{-4}\ N/C[/tex]
(b) is correct option