Join WhatsApp Icon JEE WhatsApp Group
Question 13

Two insulated circular loop $$A$$ and $$B$$ radius $$a$$ carrying a current of $$I$$ in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be:

image

$$B_1 = B_2 = \frac{\mu_0 I}{2a}$$

$$B_{\text{net}} = \sqrt{B_1^2 + B_2^2}$$ ($$B_1$$ and $$B_2$$ are perpendicular vectors)

$$B_{\text{net}} = \sqrt{\left(\frac{\mu_0 I}{2a}\right)^2 + \left(\frac{\mu_0 I}{2a}\right)^2} \implies B_{\text{net}} = \sqrt{2} \cdot \frac{\mu_0 I}{2a} = \frac{\mu_0 I}{\sqrt{2}a}$$

Get AI Help

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests
Ask AI