Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical


the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero


B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)





B = 0.1 T
Angle α between the velocity of theelectron and the magnetic field of the wire will be 90 degree
Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2

Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;

T = 7.83 X10⁻⁷ s
Answer:
Option C
Explanation:
According to the question:
Force exerted by the team towards south, F = 10 N
Force exerted by the opposite team towards North, F' = 17 N
Net Force, 

Thus the force will be along the direction of force whose magnitude is higher
Therefore,
towards North
Answer:
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