Question

A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 metre. The value of the magnetic field produced at the centre will be $\left( \mu _ { 0 } - \right.$ permeability for vacuum)

• A. $\frac { 10 ^ { - 7 } } { \mu _ { 0 } }$
• B. $10 ^ { - 17 } \mu _ { 0 }$
• C. $10 ^ { - 6 } \mu _ { 0 }$
• D. $10 ^ { - 7 } \mu _ { 0 }$

Answer 1

Answer: (A)

$\frac { 10 ^ { - 7 } } { \mu _ { 0 } }$

Answer 2

Case 1 : $\mathrm { B } _ { \mathrm { A } } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { \mathrm { i } } { \mathrm { r } } \otimes$

$B _ { B } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r }$ ◉

$B _ { C } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r }$◉

So net magnetic field at the centre of case 1

$B _ { 1 } = B _ { B } - \left( B _ { A } + B _ { C } \right)$

⇒ $B _ { 1 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { \pi i } { r }$ ◉ ..... (i)

Case 2 :

As we discussed before magnetic field at the centre O in this case

$B _ { 2 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { \pi i } { r } \otimes$ .....(ii)

Case 3 : $B _ { A } = 0$

$B _ { B } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { ( 2 \pi - \pi / 2 ) } { r } \otimes$ $= \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { 3 \pi i } { 2 r } \otimes$

$B _ { C } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r }$ ◉

So net magnetic field at the centre of case 3

$B _ { 3 } = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { i } { r } \left( \frac { 3 \pi } { 2 } - 1 \right) \otimes$ ....(iii)

From equation (i), (ii) and (iii) $= - \frac { \pi } { 2 } : \frac { \pi } { 2 } : \left( \frac { 3 \pi } { 4 } - \frac { 1 } { 2 } \right)$