CBSE BOARD XII, asked by trivsidd32, 9 months ago

3. Figure (34-E17) shows a convex lens of focal length
12 cm lying in a uniform magnetic field B of magnitude
1.2 T parallel to its principal axis. A particle having a
charge 2:0 x 10-C and mass 2:0 x 10 - kg is projected
perpendicular to the plane of the diagram with a speed
of 4:8 ms. The particle moves along a circle with its
centre on the principal axis at a distance of 18 cm from
the lens. Show that the image of the particle goes along
a circle and find the radius of that circle.​

Answers

Answered by Hemalathajothimani
8

Answer:

Explanation:

ANSWER

According to the given criteria,

Focal length of the convex lens, f=12cm

Uniform magnetic field, B=1.2T

Charge on the particle, q=2.0×10  

−3

C

Mass of the particle, m=2.0×10  

−6

kg

Speed of the particle, v=4.8m/s

The distance of the particle from the lens, object distance, u=−18cm

As particle is projected perpendicular to the plane.

Let the radius of the circle on which the object is moving be r

We know,

r=  

qB

mv

​  

 

⟹r=  

2.0×10  

−3

×1.2

2.0×10  

−6

×4.8

​  

 

⟹r=4×10  

−3

m=0.4cm

Now applying the lens formula, we get

f

1

​  

=  

v

1

​  

−  

u

1

​  

 

⟹  

12

1

​  

=  

v

1

​  

−  

(−18)

1

​  

 

⟹  

v

1

​  

=  

12

1

​  

−  

18

1

​  

 

⟹  

v

1

​  

=  

36

3−2

​  

 

⟹v=36cm

Let the radius of the circular path of image be r  

 

So magnification, m=−  

u

v

​  

=  

r

r  

 

​  

 

⟹−  

(−18)

36

​  

=  

0.4

r  

 

​  

 

⟹r  

=  

18

36×0.4

​  

 

⟹r  

=0.8cm

So the radius of the circular path in which the image movesis 0.8 cm.

Similar questions
Math, 1 year ago