Q.42. A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying

Q: What is the correct answer for JEE Main Physics Q42?

The correct answer is A. The magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2R}$. Given $I = 2$ A and $R = 10$ cm $= 0.1$ m. So, $B = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6}$ T. The volume of the cube is $V = (1 \text{ mm})^3 = (10^{-3} \text{ m})^3 = 10^{-9}$ m$^3$. The magnetic energy stored is $U = \frac{B^2}{2\mu_0} V = \frac{(4\pi \times 10^{-6})^2}{2 \times 4\p

Q: Q.42. 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 J. The value of is ( Tm/A, )

The correct answer is A. The magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2R}$. Given $I = 2$ A and $R = 10$ cm $= 0.1$ m. So, $B = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6}$ T. The volume of the cube is $V = (1 \text{ mm})^3 = (10^{-3} \text{ m})^3 = 10^{-9}$ m$^3$. The magnetic energy stored is $U = \frac{B^2}{2\mu_0} V = \frac{(4\pi \times 10^{-6})^2}{2 \times 4\p

Q: What are the options for this JEE Main Physics question?

A. 28 · B. 28 \times 10^{-6} · C. 628 · D. 28 \times 10^4

Question

Q.42. 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  J. The value of  is ( Tm/A, )

FirstInTest · Accepted Answer

The magnetic field at the center of a circular loop is $B = \frac{\mu_0 I}{2R}$. Given $I = 2$ A and $R = 10$ cm $= 0.1$ m. So, $B = \frac{4\pi 	imes 10^{-7} 	imes 2}{2 	imes 0.1} = 4\pi 	imes 10^{-6}$ T. The volume of the cube is $V = (1 	ext{ mm})^3 = (10^{-3} 	ext{ m})^3 = 10^{-9}$ m$^3$. The magnetic energy stored is $U = \frac{B^2}{2\mu_0} V = \frac{(4\pi 	imes 10^{-6})^2}{2 	imes 4\pi 	imes 10^{-7}} 	imes 10^{-9} = \frac{16\pi^2 	imes 10^{-12}}{8\pi 	imes 10^{-7}} 	imes 10^{-9} = 2\pi 	imes 10^{-14}$ J. Given $\pi = 3.14$, $U = 2 	imes 3.14 	imes 10^{-14} = 6.28 	imes 10^{-14}$ J. Thus, $\alpha = 6.28$.

Answer

28

Answer

28 	imes 10^{-6}

Answer

628

Answer

28 	imes 10^4

Related Questionsin Physics

More from JEE Main

Browse Engineering Questions by Subject

Practice More JEE Main Questions