### Cosmic-Ray Proton in Interstellar SpaceA cosmic-ray proton in interstellar space has an energy of **12.5 MeV** and executes a circular orbit with a radius equal to that of Mars' orbit around the Sun (**2.28 × 10¹¹ m**). What is the magnetic field (in T) in that region of space?**Hint:**\[ F_c = F_B \]\[\frac{mv^2}{r} = qvB \quad \text{this will give you} \quad B = \frac{mv}{qr} = \frac{p}{qr}\]Now think:\[K.E = \frac{1}{2}mv^2\]Where K.E can be expressed in terms of Momentum (mv) where \( p \) is momentum, and K.E is given in the question, but you need to convert that into Joules.\[(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J})\]Mass of a proton is \( 1.67 \times 10^{-27} \, \text{Kg} \).Hope you can do this with this hint!

When a charged particle travels in a magnetic field with some velocity it experience a magnetic…

A cosmic-ray proton in interstellar space has an energy of 12.5 MeV and executes a circular orbit having a radius equal to that of Mars' orbit around the Sun (2.28 × 1011 m). What is the magnetic field (in T) in that region of space? Hint: Fc =FB mv2 = gvB this will give you B = r mv = : now think K.E = gr where K.E can be expressed in terms of Momentum (mv) where p qr and K.E is given in the question but you need to convert that into the Joule (1 eV = 1.6 × 10-19 /) and mass of proton is 1.67× 10¬27 Kg Hope you can do this with this hint

A cosmic-ray proton in interstellar space has an energy of 12.5 MeV and executes a circular orbit having a radius equal to that of Mars' orbit around the Sun (2.28 × 1011 m). What is the magnetic field (in T) in that region of space? Hint: Fc =FB mv2 = gvB this will give you B = r mv = : now think K.E = gr where K.E can be expressed in terms of Momentum (mv) where p qr and K.E is given in the question but you need to convert that into the Joule (1 eV = 1.6 × 10-19 /) and mass of proton is 1.67× 10¬27 Kg Hope you can do this with this hint