# GATE 2015 EE – SET 1 Question -12

GATE 2015 EE – SET 1 Question -12,Answer Key & Full Solutions to GATE 2017 Questions Papers for All branches .Full Solutions to GATE 2017 Questions Papers ## GATE 2015 EE – SET 1 Question -12 search engine Keywords Thesis 123

• GATE 2015 EE – SET 1 Question -12
• gate paper
• gate question paper
• gate 2017 question paper with solution
• gate previous papers
• gate solved papers/li>
• gate 2017 question paper
• gate question bank
• gate question paper with solution
• gate electrical engineering
• gate model papers
• gate sample papers
• gate 2017 key
• gate previous year papers
• gate papers for electronics and communication
• gate previous year question papers with solutions for ece
• gate paper for mechanical engineering
• gate online test
• gate previous year question papers with solutions
• gate solved question papers
• gate mechanical solved papers
• gate previous year question papers with solutions for cse
• gate test papers
• gate civil engineering solved papers pdf
• gate online coaching
• gate solved papers for mechanical engineering
• made easy gate previous papers
• gate question bank for mechanical engineering
• gate previous papers with solutions for mechanical
• gate ece solved papers
• gate question bank for mechanical
• gate civil previous papers with solutions
• made easy solved paper of gate
• gate engineering mathematics previous years question papers with solutions
• gate previous papers with solutions for cse
• gate ee previous papers
• gate previous year papers for mechanical
• gate previous year question papers with solutions for engineers
• gate 10 years solved papers

Solution :

$$\vec{B}$$=$$\mu _{0}\vec{H}$$

H = $$H_{1} + H_{2}$$   , where H =$$\frac{I}2{\pi }R$$

H =$$\frac{I}{2\pi R_{1}} + \frac{I}{2\pi R_{2}}$$

H =  $$\frac{I}{2\pi \left [ \frac{3L}{2} \right ]} + \frac{I}{2\pi \left [ \frac{L}{2} \right ]}$$

H = $$\frac{I}{2\pi }\left [ \frac{2}{3L} + \frac{2}{L} \right ]$$

H = $$\frac{I}{2\pi }\left [ \frac{6+2}{3L} \right ]$$

H = $$\frac{8I}{\pi 6L}$$ = $$\frac{4I}{\pi 3L}$$ (-az).

$$\vec{B}$$=$$\mu _{0}\vec{H}$$ = $$\frac{4\mu_0 I}{3\pi L}(-a_{z})$$ =  $$\frac{-4\mu 0I}{3\pi L}$$