Answer:
The time taken will be 2.73*10^(-8) sec
Explanation:
After the proton enters in to a magnetic field, it follows circular motion in a clockwise direction.
Proton will re-emerge into the field free region, when it has completed semi circle.
Let the time required be t sec
Force = qv*B
centripetal force= mv^2/r
Now taking both the equations
r = mv/ qB = 8.7*10^(-3) m
Total distance traveled will be= pi*r
And, total time taken will be t = pi*r / v = 2.73*10^(-8) sec
Answer:
The time required for the proton to re-emerge into the field-free region is 5.47 x 10⁻⁸ s
Explanation:
given;
mass of proton, m = 1.67 × 10⁻²⁷ kg
charge of proton, q = 1.6 × 10⁻¹⁹ C
speed of proton, v = 1 × 10⁶ m/s
uniform magnetic field, B = 1.2 T
Force on the proton due to uniform magnetic field = qvB
Force on the proton in the circular region = ma = mv²/r
The time required for the proton to re-emerge into the field-free region, is the same as the time required for the proton to return to its starting position. During this period, the proton has made one complete revolution.
v = 2πr/t
t = 2πr/v
where;
t is the required for the proton to re-emerge into the field-free region
r is the radius of the circular region
mv²/r = qvB
r = mv²/qvB
r = mv/qB
r = (1.67 × 10⁻²⁷ x 1 × 10⁶) / (1.6 × 10⁻¹⁹ x 1.2)
r = 8.698 x 10⁻³ m
t = 2πr/v
t = (2π x 8.698 x 10⁻³) / (1 × 10⁶)
t = 5.47 x 10⁻⁸ s
Therefore, the time required for the proton to re-emerge into the field-free region is 5.47 x 10⁻⁸ s
