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  • 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. Mag

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 O is :

Option: 1

\overrightarrow{\mathrm{B}}=-\frac{\mu_0}{4 \pi} \frac{1}{\mathrm{R}}(\mu \hat{\mathrm{i}} \times 2 \hat{\mathrm{k}})


Option: 2

\overrightarrow{\mathrm{B}}=-\frac{\mu_0}{4 \pi} \frac{1}{\mathrm{R}}(\pi \hat{\mathrm{i}}+2 \hat{\mathrm{k}})


Option: 3

\overrightarrow{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{1}{\mathrm{R}}(\pi \hat{\mathrm{i}}-2 \hat{\mathrm{k}})


Option: 4

\overrightarrow{\mathrm{B}}=\frac{\mu_0}{4 \pi} \frac{1}{\mathrm{R}}(\pi \hat{\mathrm{i}}+2 \hat{\mathrm{k}})


Answers (1)

best_answer

Magnetic field due to segment ' 1 '
\overrightarrow{\mathrm{B}_1}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{R}}\left[\sin 90^{\circ}+\sin 0^{\circ}\right](-\hat{\mathrm{k}}) \\
=\frac{-\mu_0 \mathrm{I}}{4 \pi \mathrm{R}}(\hat{\mathrm{k}})=\overrightarrow{\mathrm{B}_3}

Magnetic field due to segment 2
\mathrm{B}_2=\frac{\mu_0 \mathrm{I}}{4 \mathrm{R}}(-\hat{\mathrm{i}})=\frac{-\mu_0 \mathrm{I}}{4 \pi \mathrm{R}}(\pi \hat{\mathrm{r}}).

\therefore \overrightarrow{\mathrm{B}}  at centre

\overrightarrow{\mathrm{B}_{\mathrm{c}}}=\overrightarrow{\mathrm{B}_1}+\overrightarrow{\mathrm{B}_2}+\overrightarrow{\mathrm{B}_3}=\frac{-\mu_0 \mathrm{I}}{4 \pi \mathrm{R}}(\pi \hat{\imath}+2 \mathrm{k})
 

Posted by

SANGALDEEP SINGH

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