Answer:
The appropriate solution is:
(a) [tex]\frac{1}{4}(I_o)[/tex]
(b) [tex]\frac{1}{4} (u_o)[/tex]
(c) [tex]\frac{1}{2}B_o[/tex]
Explanation:
According to the question, the value is:
Power of bulb,
= 60 W
Distance,
= 1.0 mm
Now,
(a)
⇒ [tex]\frac{I}{I_o} =\frac{r_o_2}{r_2}[/tex]
On applying cross-multiplication, we get
⇒ [tex]I=I_o\times \frac{1_2}{2^2}[/tex]
⇒ [tex]=I_o\times \frac{1}{4}[/tex]
⇒ [tex]=\frac{1}{4} (I_o)[/tex]
(b)
As we know,
⇒ [tex]\frac{u}{u_o} =\frac{I}{I_o}[/tex]
By putting the values, we get
⇒ [tex]u=\frac{1}{4}(u_o)[/tex]
(c)
⇒ [tex]\frac{B^2}{B_o^2} =\frac{u}{u_o}[/tex]
[tex]=\frac{I}{I_o}[/tex]
⇒ [tex]B=B_o\times \sqrt{\frac{1}{4} }[/tex]
⇒ [tex]=\frac{1}{2}(B_o)[/tex]
