Answer:
the magnitude of the magnetic field is 1.23 x 10⁸ T.
Explanation:
Given;
magnitude of the electric field, E = 3.7 V/m
The magnitude of the magnetic field is calculated as;
E = cB
where;
B is the magnitude of the magnetic field
c is the speed of light = 3 x 10⁸ m/s
From the above equation, the magnetic field, B, is calculated as;
Therefore, the magnitude of the magnetic field is 1.23 x 10⁸ T.
The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
Speed of the proton = 5.02 × 10 ⁶ m /a
Angel of between the velocity and the magnetic force = 60 °
The magnitude of magnetic field B = 0.180 T
The magnitude of the magnetic force on the proton is,
Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
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Answer:
The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.
Explanation:
ω1= 1.72x10^4 rad/sec
ω2= 5.42x10^4 rad/sec
ωmax= 8.42x10^4 rad/sec
θ= 1.72x10^4 rad
α=7.67 x10^4 rad/sec²
t= ωmax / α
t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²
t=1.097 sec
Answer:
so here it will move in circle with radius 4.06 cm
Explanation:
As we know that proton is moving towards west while the magnetic field is vertically upwards
So here the force on the proton must be perpendicular to the velocity
So here we have
so here we have
since force is perpendicular to the velocity so here it must be centripetal force
here we have
so we have
so here it will move in circle with radius 4.06 cm