Q.49. A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a

Q: What is the correct answer for JEE Main Physics Q49?

The correct answer is 314. Sol. The torque acting on a current loop in a magnetic field is given by $\vec{\tau} = \vec{M} \times \vec{B}$, where $\vec{M}$ is the magnetic dipole moment of the coil. The magnitude of the torque is $\tau = MB \sin \theta$. The magnetic dipole moment of a circular coil is $M = NIA$, where N is the number of turns, I is the current, and A is the area of the coil. Given: $N = 125$, $I = 1 \text{

Q: Q.49. A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of with the…

The correct answer is 314. Sol. The torque acting on a current loop in a magnetic field is given by $\vec{\tau} = \vec{M} \times \vec{B}$, where $\vec{M}$ is the magnetic dipole moment of the coil. The magnitude of the torque is $\tau = MB \sin \theta$. The magnetic dipole moment of a circular coil is $M = NIA$, where N is the number of turns, I is the current, and A is the area of the coil. Given: $N = 125$, $I = 1 \text{

Question

Q.49. A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of  with the direction of the magnetic field. The torque acting on the coil is . The value of  is __________ ()

FirstInTest · Accepted Answer

Sol. The torque acting on a current loop in a magnetic field is given by $\vec{	au} = \vec{M} 	imes \vec{B}$, where $\vec{M}$ is the magnetic dipole moment of the coil. The magnitude of the torque is $	au = MB \sin 	heta$. The magnetic dipole moment of a circular coil is $M = NIA$, where N is the number of turns, I is the current, and A is the area of the coil. Given: $N = 125$, $I = 1 	ext{ A}$, $r = 2 	ext{ cm} = 0.02 	ext{ m}$. So, $A = \pi r^2 = \pi (0.02)^2 = 4\pi 	imes 10^{-4} 	ext{ m}^2$. $M = 125 	imes 1 	imes 4\pi 	imes 10^{-4} = 500\pi 	imes 10^{-4} = 0.05\pi 	ext{ A.m}^2$. Given $B = 0.4 	ext{ T}$ and $	heta = 30^\circ$ (angle between the axis of the coil and the magnetic field, which is the angle between $\vec{M}$ and $\vec{B}$). $	au = (0.05\pi) 	imes 0.4 	imes \sin(30^\circ) = (0.05\pi) 	imes 0.4 	imes 0.5 = 0.01\pi 	ext{ N.m}$. We need $	au = a 	imes 10^{-4} 	ext{ N.m}$. So, $0.01\pi = a 	imes 10^{-4}$. $a = \frac{0.01\pi}{10^{-4}} = \frac{\pi 	imes 10^{-2}}{10^{-4}} = \pi 	imes 10^2 = 100\pi$. Using $\pi = 3.14$, $a = 100 	imes 3.14 = 314$. The solution's calculation is $	au = 125 	imes 1 	imes \pi (\frac{2}{100})^2 	imes 0.4 	imes \sin 30^\circ = 125 	imes \pi 	imes 4 	imes 10^{-4} 	imes 0.4 	imes 0.5 = 500\pi 	imes 10^{-4} 	imes 0.2 = 100\pi 	imes 10^{-4} 	ext{ N.m}$. So $a = 100\pi = 314$. This matches the answer.

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