Magnetic Effects Of Current Question 425
Question: A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point 0 is:
Options:
A) $ \overset{\to }{\mathop{B}},=-\frac{{\mu_{0}}}{4\pi }\frac{I}{R}( \mu \hat{i}\times 2\hat{k} ) $
B) $ \overset{\to }{\mathop{B}},=-\frac{{\mu_{0}}}{4\pi }\frac{I}{R}( \pi \hat{i}+2\hat{k} ) $
C) $ \overset{\to }{\mathop{B}},=\frac{{\mu_{0}}}{4\pi }\frac{I}{R}( \pi \hat{i}-2\hat{k} ) $
D) $ \overset{\to }{\mathop{B}},=\frac{{\mu_{0}}}{4\pi }\frac{I}{R}( \pi \hat{i}+2\hat{k} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Magnetic field due to segment ‘1’ $ {{\overset{\to }{\mathop{B}}}_{1}}$=
$\frac{{\mu_{0}}I}{4\pi R}\sin 90^{o}+\sin 0^{o}(-\hat{k}) $
$ =\frac{-{\mu_{0}}I}{4\pi R}(\hat{k})={{\overset{\to }{\mathop{B}}}_{3}} $ Magnetic field due to segment 2
$ B_{2}=\frac{{\mu_{0}}I}{4R}(-\hat{i})=\frac{-{\mu_{0}}I}{4\pi R}(\pi \hat{i}) $
$ \therefore ,\overset{\to }{\mathop{B}}, $ at centre
$ {{\overset{\to }{\mathop{B}}}_{c}}$=
${{\overset{\to }{\mathop{B}}}_{1}}$
$+\overset{\to }{\mathop{B_{2}+}}\overset{\to }{\mathop{B_{3}}} $
$ =\frac{-{\mu_{0}}I}{4\pi R}(\pi \hat{i}+2\hat{k}) $
BETA
