Electrons move at right angle to a magnetic field of 1.5 × 10 − 2 Tesla with a speed of 6 × 10 7 m/s. If the specific charge of the electron is 1.7 × 10 11 C/kg, then the radius of the circular path will be? Options: A . &nbsp2.9 cm B . &nbsp3.9 cm C . &nbsp2.35 cm D . &nbsp3 cm

Answer: Option C : C r = m v q B ⇒ v ( q / m ) . B = 6 × 10 7 1.7 × 10 11 × 1.5 × 10 − 2 = 2.35 × 10 − 2 m = 2.35 c m

Electrons move at right angle to a magnetic field of 1.5×10−2 Tesla

Question

Electrons move at right angle to a magnetic field of

1.5 \times 10^{- 2}

Tesla with a speed of

6 \times 10^{7}

m/s. If the specific charge of the electron is

1.7 \times 10^{11}

C/kg, then the radius of the circular path will be?

Options:

A . 2.9 cm

B . 3.9 cm

C . 2.35 cm

D . 3 cm

Answer: Option C
:
C

r = \frac{m v}{q B} \Rightarrow \frac{v}{(q / m) . B} = \frac{6 \times 10^{7}}{1.7 \times 10^{11} \times 1.5 \times 10^{- 2}} = 2.35 \times 10^{- 2} m = 2.35 c m

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