"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 $\times$ 10–27 kg, charge of the proton = 1.69 $\times$ 10–19 C)"

Question

Accepted Answer

"Pitch = $\frac{2\pi m}{qB}$ vcos$\theta$

= $${{2\left( {3.14} \right)\left( {1.67 \times {{10}^{ - 27}}} \right) \times 4 \times {{10}^5} \times \cos 60} \over {\left( {1.69 \times {{10}^{ - 19}}} \right)\left( {0.3} \right)}}$$

= 0.04 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 $\times$ 10^–27 kg, charge
of the proton = 1.69 $\times$ 10^–19 C)

Solution

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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 $\times$ 10–27 kg, charge of the proton = 1.69 $\times$ 10–19 C)

Solution

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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 $\times$ 10^–27 kg, charge
of the proton = 1.69 $\times$ 10^–19 C)