JEE MAIN - Physics (2020 - 2nd September Morning Slot - No. 22)
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)
(Mass of the proton = 1.67 $$ \times $$ 10–27 kg, charge
of the proton = 1.69 $$ \times $$ 10–19 C)
2 cm
12 cm
5 cm
4 cm
Explanation
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
= $${{2\left( {3.14} \right)\left( {1.67 \times {{10}^{ - 27}}} \right) \times 4 \times {{10}^5} \times \cos 60} \over {\left( {1.69 \times {{10}^{ - 19}}} \right)\left( {0.3} \right)}}$$
= 0.04 m = 4 cm
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