A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y
= 3.0 m. In this region of space, the magnetic field is equal to 40 mT in the positive z direction. What is the magnitude of the force on this segment of wire?
1 answer:
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Answer:
1.312 x 10⁻¹² J/nucleon
Explanation:
mass of ¹³⁶Ba = 135.905 amu
¹³⁶Ba contain 56 proton and 80 neutron
mass of proton = 1.00728 amu
mass of neutron = 1.00867 amu
mass of ¹³⁶Ba = 56 x 1.00728 amu + 80 x 1.00867 amu
= 137.10128 amu
mass defect = 137.10128 - 135.905
= 1.19628 amu
mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg
= 1.9858 x 10⁻²⁷ Kg
speed of light = 3 x 10⁸ m/s
binding energy,
E = mass defect x c²
E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²
E = 17.87 x 10⁻¹¹ J/atom
now,
binding energy per nucleon =
= 0.1312 x 10⁻¹¹ J/nucleon
= 1.312 x 10⁻¹² J/nucleon