JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A. The magnetic energy stored inside the cube is $$\alpha \times 10^{-14}$$ J. The value of $$\alpha$$ is _______.
$$(\mu_0 = 4\pi \times 10^{-7}$$ Tm/A, $$\pi = 3.14)$$
$$B = \frac{\mu_0 I}{2R}$$
$$u_B = \frac{B^2}{2\mu_0}$$
$$U = u_B \cdot V = \frac{B^2}{2\mu_0} \cdot V$$
$$B = \frac{(4\pi \times 10^{-7}) \times 2}{2 \times 0.1} = \frac{4\pi \times 10^{-7}}{0.1} = 4\pi \times 10^{-6} \text{ T}$$
$$U = \frac{(4\pi \times 10^{-6})^2}{2 \times (4\pi \times 10^{-7})} \times 10^{-9}$$
$$U = 2 \times 3.14 \times 10^{-14} = 6.28 \times 10^{-14} \text{ J}$$
$$\alpha = 6.28$$
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
