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Figure shows a square loop ABCD with edg...

Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the centre of the loop assuming uniform wires. `

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Since the resistance in ABC and ADC are r and 2r respectively. So , current divided as `(2i)/(3)and (i)/(3)`
B at centre due to ABC `=(mu)/(4pi) (2i)/(3). (a)/(a^2) xx 4xx sqrt2`
`=(2sqrt2mu_oi)/(3pia)`
B at centre due to ADC `=(mu_o)/(4pi).(i)/(3). (a)/(a^2)xx4xx sqrt2 `.
` `=(sqrt 2mu_oi)/(3pia)`.
So net B `=(sqrt2mu_oi)/(3pia).`
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HC VERMA-MAGNETIC FIELD DUE TO CURRENT-Exercises
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