Question

a proton is accelerated by a potential difference of 8×10⁴ volts. it then enters a uniform field of B=0.707 web/metre² in a direction making an angle of 45° with the magnetic field. what will be the radius of circular path

I need solution of this ​

Answer 1

Answer:

Explanation:

For a charge particle moving in a constant magnetic field and  

v

⊥  

B

 

r

mv  

2

 

​

=qvB⇒ r=  

qB

mv

​

=  

qB

P

​

....(i)

If e is accelerated through a potential difference of 10  

4

V, then  

K.E of electron =eV

⇒  

2m

P  

2

 

​

=eV⇒ P=  

2 meV

​

....(ii)

From (i) & (ii)

⇒r=  

qB

2meV

​

 

​

 

=  

1.6×10  

−19

×0.01

2×9.1×10  

−31

×1.6×10  

−19

×10  

4

 

​

 

​

 

=  

6.4×10  

−21

 

5.39×10  

−23

 

​

m=8.4×10  

−3

m