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A long wire bent as shown in fig. carries current I. If the radius of the semicircular portion is a, the magnetic field at centre C is A. `(mu_0I)/(4a)`B. `(mu_0I)/(4pia)sqrt(pi^2+4)`C. `(mu_0I)/(4a)+(mu_0I)/(4pia)`D. `(mu_0I)/(4pia)sqrt((pi^2-4))` |
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Answer» Correct Answer - B (b) `B_1=(mu_0)/(4pi)xx(2piI)/axx1/2 ("due to semicircular part")` `B_2=(mu_0)/(4pi)xx(2I)/a ("due to parallel parts of currents")` These two fields are at right angles to each other. Hence, resultant field `B=sqrt(B_1^2+B_2^2)=(mu_0I)/(4pia)sqrt(pi^2+4)` |
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