A beam of protons with speed 4 × 10⁵ ms⁻¹ enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton = 1.67 × 10⁻²⁷ kg, charge of the proton = 1.69 × 10⁻¹⁹ C)

Question

Accepted Answer

4 cm. T = 2*pi*(1.67e-27)/(1.69e-19*0.3) = 2.07e-7 s. Pitch = v*cos60*T = 4e5*0.5*2.07e-7 = 0.041 m ~ 4 cm.

A beam of protons with speed 4 × 10⁵ ms⁻¹ enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to: (Mass of the proton = 1.67 × 10⁻²⁷ kg, charge of the proton = 1.69 × 10⁻¹⁹ C)

Solution

Related JEE Main Physics questions