Q.50. A circular loop of radius 20 cm and resistance is placed in a time varying magnetic field T.

Q: What is the correct answer for JEE Main Physics Q50?

The correct answer is 44. Given: Radius of the loop $r = 20$ cm = $0.2$ m Resistance of the loop $R = 2\Omega$ Magnetic field $B = (2t^2+2t+3)$ T Time $t = 3$ s The magnetic flux $\Phi$ through the loop is given by $\Phi = B A \cos\theta$. Since the plane of the loop is perpendicular to the magnetic field, $\theta = 0^\circ$, so $\cos\theta = 1$. Area of the loop $A = \pi r^2 = \pi (0.2)^2 = 0.04\pi$ m². So, $\Phi = (2t^2

Q: Q.50. A circular loop of radius 20 cm and resistance is placed in a time varying magnetic field T. At t = 0, for the plane of the loop being perpendicular to the magnetic field and, the induced…

The correct answer is 44. Given: Radius of the loop $r = 20$ cm = $0.2$ m Resistance of the loop $R = 2\Omega$ Magnetic field $B = (2t^2+2t+3)$ T Time $t = 3$ s The magnetic flux $\Phi$ through the loop is given by $\Phi = B A \cos\theta$. Since the plane of the loop is perpendicular to the magnetic field, $\theta = 0^\circ$, so $\cos\theta = 1$. Area of the loop $A = \pi r^2 = \pi (0.2)^2 = 0.04\pi$ m². So, $\Phi = (2t^2

Question

Q.50. A circular loop of radius 20 cm and resistance  is placed in a time varying magnetic field T. At t = 0, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at t = 3 s is  A. The value of  is . (Take )

FirstInTest · Accepted Answer

Given:
Radius of the loop $r = 20$ cm = $0.2$ m
Resistance of the loop $R = 2\Omega$
Magnetic field $B = (2t^2+2t+3)$ T
Time $t = 3$ s

The magnetic flux $\Phi$ through the loop is given by $\Phi = B A \cos	heta$. Since the plane of the loop is perpendicular to the magnetic field, $	heta = 0^\circ$, so $\cos	heta = 1$.
Area of the loop $A = \pi r^2 = \pi (0.2)^2 = 0.04\pi$ m².
So, $\Phi = (2t^2+2t+3) 	imes 0.04\pi$.

According to Faraday's law of electromagnetic induction, the induced emf $\varepsilon = -\frac{d\Phi}{dt}$.
$\frac{d\Phi}{dt} = \frac{d}{dt} [(2t^2+2t+3) 	imes 0.04\pi] = 0.04\pi \frac{d}{dt} (2t^2+2t+3) = 0.04\pi (4t+2)$.

At $t = 3$ s:
$\frac{d\Phi}{dt} = 0.04\pi (4(3)+2) = 0.04\pi (12+2) = 0.04\pi (14) = 0.56\pi$ Wb/s.

The magnitude of the induced emf is $|\varepsilon| = 0.56\pi$ V.

The induced current $I = \frac{|\varepsilon|}{R} = \frac{0.56\pi}{2} = 0.28\pi$ A.

We are given that the induced current is $\frac{\alpha}{50}$ A.
So, $\frac{\alpha}{50} = 0.28\pi$.
Using $\pi = 22/7$:
$\frac{\alpha}{50} = 0.28 	imes \frac{22}{7} = \frac{28}{100} 	imes \frac{22}{7} = \frac{4}{100} 	imes 22 = \frac{1}{25} 	imes 22 = \frac{22}{25}$.

$\alpha = 50 	imes \frac{22}{25} = 2 	imes 22 = 44$.
Thus, $\alpha = 44$.

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