The radius of circular path of an electron when subjected to a perpendicular magnetic field, will be

Question

The radius of circular path of an electron when subjected to a perpendicular magnetic field, will be (A)  (mv/Be)  (B)  (me/B)  (C)  (mE/B)  (D)  (Be/mv)

Tardigrade Admin · Accepted Answer

Centripetal force on the electron =(mv2/r) Force on moving charge in the magnetic field is =Bev Hence (mv2/r)=Bev or r=(mv/Be)

Q. The radius of circular path of an electron when subjected to a perpendicular magnetic field, will be

$\frac{m v}{B e}$

$\frac{m e}{B}$

$\frac{m E}{B}$

$\frac{B e}{m v}$

Centripetal force on the electron $= \frac{m v ^{2}}{r}$ Force on moving charge in the magnetic field is $= B e v$ Hence $\frac{m v ^{2}}{r} = B e v$ or $r = \frac{m v}{B e}$

Q. The radius of circular path of an electron when subjected to a perpendicular magnetic field, will be

Bemv​

Bme​

BmE​

mvBe​

Centripetal force on the electron =rmv2​ Force on moving charge in the magnetic field is =Bev Hence rmv2​=Bev or r=Bemv​

$\frac{m v}{B e}$

$\frac{m e}{B}$

$\frac{m E}{B}$

$\frac{B e}{m v}$

Centripetal force on the electron $= \frac{m v ^{2}}{r}$ Force on moving charge in the magnetic field is $= B e v$ Hence $\frac{m v ^{2}}{r} = B e v$ or $r = \frac{m v}{B e}$