Question

For the solenoid shown in figure. The magnetic field at point P is

• A. $\frac { \mu _ { 0 } \mathrm { ni } } { 4 } ( \sqrt { 3 } + 1 )$
• B. $\frac { \sqrt { 3 } \mu _ { 0 } \mathrm { ni } } { 4 }$
• C.
• D. $\frac { \mu _ { 0 } \mathrm { ni } } { 4 } ( \sqrt { 3 } - 1 )$

Answer 1

Answer: (A)

$\frac { \mu _ { 0 } \mathrm { ni } } { 4 } ( \sqrt { 3 } + 1 )$

Answer 2

$B = \frac { \mu _ { 0 } } { 4 \pi } \cdot 2 \pi n i ( \sin \alpha + \sin \beta )$ . From figure α = (90^o – 30^o)

= 60^o and β = (90^o – 60^o) = 30^o

∴ $B = \frac { \mu _ { 0 } n i } { 2 } \left( \sin 60 ^ { \circ } + \sin 30 ^ { \circ } \right) = \frac { \mu _ { 0 } n i } { 4 } ( \sqrt { 3 } + 1 )$ .