Q.An electron moving with speed $10^6$ m/s enters a magnetic field of 0.2 T at $60^\circ$ . The pitch of the helical path is $(e = 1.6 × 10^{-19} C, m_e = 9.1 × 10^{-31}$ kg)

2.8 \times 10^{-3}

5.6 \times 10^{-3}

8.4 \times 10^{-3}

11.2 \times 10^{-3}

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.An electron moving with speed 10610^6106 m/s enters a magnetic field of 0.2 T at 60∘60^\circ60∘. The pitch of the helical path is (e=1.6×10−19C,me=9.1×10−31(e = 1.6 × 10^{-19} C, m_e = 9.1 × 10^{-31}(e=1.6×10−19C,me​=9.1×10−31 kg)