Question

Use the following information to answer the next question The element un unquadium (11% Uhug) has been created by fusing calcium ions Ca) with plutonium nuclei (24.4 Pu). The cakium ions are doubly charged (+2e) and have a mass of 7.96 x 10-26 kg. To accelerate these ions to a high enough energy to fuse with plutonium, they are repeatedly accelerated by an electric potential difference. They are contained in a magnetic field betweea these accelerations. In one stage of the acceleration process, calcium ions enter the accelerating chamber at a speed of 1.00 x 10 m/s and exit it at a speed of 275 x 106 m/s. They immediately enter a magnetic field and follow a path that has a radius of 1.24 m. Numerical Response 2. The strength of the magnetic field used to contain the calcium ions, expressed in scientific notation, is abc x 10-41. The values of a, b, c, and d and are (Record all four digits of your answer in the numerical-response section on the answer sheet.)

Answer 1

Charge on ion q= 2e = 2* 1.6 * 10^-19 C

Mass of ion m= 7.96*10^-26 kg

Now, the ions after emerging from the electric field , enter a region of magnetc field with a velocity 2.75*10⁶ m/s.

The magnetic Force on the ion balances the centripetal force and accounts for the rotational motion of the ion.

Magnetic force on the ion = qvB

Centripetal force = mv²/r

Thus, qvB=mv²/r

Thus, the radius of path

$r=\frac{mv}{qB}=>1.24= \frac{7.96 \times 10^{-26} \times 2.75 \times 10^{6}}{2\times 1.6 \times 10^{-19} \times B}$

Thus, B=0.55T       (strength of magnetic field)

So, a=0,b=5,c=5,d=0