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$B = 3[\frac{\mu_0 I (sin\alpha + sin\beta)}{4\pi r}$] $\text{here } \alpha = \beta = 60^o$ $r = \frac{a}{2\sqrt 3}$ $\Rightarrow B = 3[\frac{\mu_0 I (sin 60^o + sin 60^o)}{4\pi r}] = 3[\frac{\mu_0 I \sqrt 3}{4\pi r}] = \frac{9\mu_0 I }{2\pi a}$ |
