A uniform magnetic field is at right angle to the direction of motion of proton. As a result, the proton describes a circular path of radius 2.5 cm. If the speed of proton is doubled then the radius of the circular path will be :

Question

A uniform magnetic field is at right angle to the direction of motion of proton. As a result, the proton describes a circular path of radius 2.5 cm. If the speed of proton is doubled then the radius of the circular path will be : (A) 0.5 cm (B) 2.5 cm (C) 5.0 cm (D) 7.5 cm

Tardigrade Admin · Accepted Answer

(mv2/r)=evB r=(mv/eB) (r2/r1)=(v2/v1) (r2/2.5)=(2v/v) ⇒ r2=5 cm

Q. A uniform magnetic field is at right angle to the direction of motion of proton. As a result, the proton describes a circular path of radius 2.5 cm. If the speed of proton is doubled then the radius of the circular path will be :

0.5 cm

2.5 cm

5.0 cm

7.5 cm

$\frac{m v ^{2}}{r} = e v B$ $r = \frac{m v}{e B}$ $\frac{r _{2}}{r _{1}} = \frac{v _{2}}{v _{1}}$ $\frac{r _{2}}{2.5} = \frac{2 v}{v}$ $\Rightarrow$ $r_{2} = 5 c m$

Q. A uniform magnetic field is at right angle to the direction of motion of proton. As a result, the proton describes a circular path of radius 2.5 cm. If the speed of proton is doubled then the radius of the circular path will be :

0.5 cm

2.5 cm

5.0 cm

7.5 cm

rmv2​=evB r=eBmv​ r1​r2​​=v1​v2​​ 2.5r2​​=v2v​ ⇒ r2​=5cm

$\frac{m v ^{2}}{r} = e v B$ $r = \frac{m v}{e B}$ $\frac{r _{2}}{r _{1}} = \frac{v _{2}}{v _{1}}$ $\frac{r _{2}}{2.5} = \frac{2 v}{v}$ $\Rightarrow$ $r_{2} = 5 c m$