A proton beam enters a magnetic field of 10-4 Wb m-2 normally. If the specific charge of the proton is 1011 C kg-1 and its velocity is 109 m s-1 then the radius of the ?circle decribed will be

Question

A proton beam enters a magnetic field of 10-4 Wb m-2 normally. If the specific charge of the proton is 1011 C kg-1 and its velocity is 109 m s-1 then the radius of the ?circle decribed will be (A) 0.1 m (B) 10 m (C) 100 m (D) 1 m

Tardigrade Admin · Accepted Answer

Given, β=10-4 Wb m -2 (q/m) =1011 C kg -1 v =109 ms -1 We know that, radius of the circle, I =(m v/q B) I =(v/((q)m) B) r =(109/(1011)(10-4)) I =100 m

Q. A proton beam enters a magnetic field of $1 0^{- 4} Wb m^{- 2}$ normally. If the specific charge of the proton is $1 0^{11} C k g^{- 1}$ and its velocity is $1 0^{9} m s^{- 1}$ then the radius of the ?circle decribed will be

0.1 m

10 m

100 m

1 m

Given, $β = 1 0^{- 4} Wb m^{- 2}$
$\frac{q}{m} = 1 0^{11} C k g^{- 1}$
$v = 1 0^{9} m s^{- 1}$
We know that, radius of the circle,
$I = \frac{m v}{qB}$
$I = \frac{v}{( \frac{q}{m} ) B}$
$r = \frac{1 0 ^{9}}{( 1 0 ^{11} ) ( 1 0 ^{- 4} )}$
$I = 100 m$

Q. A proton beam enters a magnetic field of 10−4Wbm−2 normally. If the specific charge of the proton is 1011Ckg−1 and its velocity is 109ms−1 then the radius of the ?circle decribed will be

0.1 m

10 m

100 m

1 m

Given, β=10−4Wbm−2 mq​=1011Ckg−1 v=109ms−1 We know that, radius of the circle, I=qBmv​ I=(mq​)Bv​ r=(1011)(10−4)109​ I=100m

Q. A proton beam enters a magnetic field of $1 0^{- 4} Wb m^{- 2}$ normally. If the specific charge of the proton is $1 0^{11} C k g^{- 1}$ and its velocity is $1 0^{9} m s^{- 1}$ then the radius of the ?circle decribed will be

Given, $β = 1 0^{- 4} Wb m^{- 2}$
$\frac{q}{m} = 1 0^{11} C k g^{- 1}$
$v = 1 0^{9} m s^{- 1}$
We know that, radius of the circle,
$I = \frac{m v}{qB}$
$I = \frac{v}{( \frac{q}{m} ) B}$
$r = \frac{1 0 ^{9}}{( 1 0 ^{11} ) ( 1 0 ^{- 4} )}$
$I = 100 m$