An electron having momentum 2.4 × 10-23 kg ms -1 enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of 30° with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be

Question

An electron having momentum 2.4 × 10-23 kg ms -1 enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of 30° with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be (A) 2 mm (B) 1 mm (C) (√3/2) mm (D) 0.5 mm

Tardigrade Admin · Accepted Answer

The radius of the helical path of the electron in the uniform magnetic field is

r = \frac{m v _{⊥}}{e B}

= \frac{m v s i n θ}{e B}

= \frac{( 2.4 \times 1 0 ^{- 23} k g m s ^{- 1} ) \times s i n 3 0 ^{\circ}}{( 1.6 \times 1 0 ^{- 19} C ) \times ( 0.15 T )}

= 5 \times 1 0^{- 4} m

= 0.5 \times 1 0^{- 3} m

= 0.5 mm

Q. An electron having momentum $2.4 \times 1 0^{- 23} k g m s^{- 1}$ enters a region of uniform magnetic field of $0.15 T$ . The field vector makes an angle of $3 0^{\circ}$ with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be

$2 mm$

$1 mm$

$\frac{3}{2} mm$

$0.5 mm$

Q. An electron having momentum 2.4×10−23kgms−1 enters a region of uniform magnetic field of 0.15T. The field vector makes an angle of 30∘ with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be

2mm

1mm

23​​mm

0.5mm

The radius of the helical path of the electron in the uniform magnetic field is r=eBmv⊥​​ =eBmvsinθ​ =(1.6×10−19C)×(0.15T)(2.4×10−23kgms−1)×sin30∘​ =5×10−4m =0.5×10−3m =0.5mm

Q. An electron having momentum $2.4 \times 1 0^{- 23} k g m s^{- 1}$ enters a region of uniform magnetic field of $0.15 T$ . The field vector makes an angle of $3 0^{\circ}$ with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be

$2 mm$

$1 mm$

$\frac{3}{2} mm$

$0.5 mm$