First slide

Magnetic field due to a current carrying Element - Biot - Savart law

Question

is a straight wire having current 2I, and ABDE coplanar with PQ is the combination of wires and circular arc having current I. If BC is $\frac{R}{2}$ and C is at distance R from PQ, then magnetic field at C is

Difficult

A

$\frac{μ_{0} I}{R} [\frac{1}{3} + \frac{1}{4 π}] o u t w a r d$
B

$\frac{μ_{0} I}{4 R} i n w a r d$
C

$\frac{μ_{0} I}{4 R} (1 + \frac{4}{π}) i n w a r d$
D

$\frac{μ_{0}}{4 R} (\frac{1}{3} + \frac{4}{π}) o u t w a r d$

Solution

$f o r t h e p a r t A B a n d D E m a g n e t i c f i e l d i s z e r o a s t h e y a r e o n t h e s a m e l i n e d u e t o a r c B D i t i s \frac{(1}{12}) t h o f a c i r c l e o f r a d i u s \frac{R}{2} a n d f o r t h e s t r a i g h t c o n d u c t i o r i t i s a t a d i s \tan c e R . s o t o t a l m a g n e t i c f i e l d i s \frac{1}{12} (\frac{μ_{0} I}{2 \frac{R}{2}}) a n d \frac{μ_{0} 2 I}{2 π R} a n d b o t h a r e o u t w a r d$

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