Final answer:
The strength of the magnetic field in which a 7.35\(\times\)10\(^{\-26}\) kg molecule with a charge of +2e, traveling at a speed of 23.4 m/s moves in a circular arc of radius 0.0706 m, is 7.2\(\times\)10\(^{\-4}\) Tesla.
Step-by-step explanation:
To find the magnetic field in which a charged particle moves in a circular path, we can use the formula that relates the magnetic force to the centripetal force required to maintain the circular motion:
\( F_B = F_c \)
where \( F_B \) is the magnetic force and \( F_c \) is the centripetal force.
The magnetic force can be calculated by:
\( F_B = qvB \)
and the centripetal force is given by:
\( F_c = \frac{mv^2}{r} \)
By setting these two equations equal to each other, and solving for B (the magnetic field strength), we get:
\( B = \frac{mv}{qr} \)
Given the mass (m) of the molecule as 7.35\(\times\)10\(^{\-26}\) kg, charge (q) as 2e or 2\(\times\)3.20\(\times\)10\(^{\-19}\) C, velocity (v) as 23.4 m/s, and radius (r) of the circular path as 0.0706 m,
\( B = \frac{(7.35\(\times\)10\(^{\-26}\) kg)(23.4 m/s)}{(2\(\times\)3.20\(\times\)10\(^{\-19}\) C)(0.0706 m)} = 7.2 \times 10\(^{\-4}\) T \)
So the strength of the magnetic field is 7.2\(\times\)10\(^{\-4}\) Tesla.