A square loop of side 10 cm and resistance 0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given:Side of square loop,$l = 10 \ cm = 0.1 \ m$Area of loop,$A = l^2 = (0.1 \ m)^2 = 0.01 \ m^2 = 100 \ cm^2$Resistance,$R = 0.5 \ \Omega$Initia

Vedclass · Accepted Answer

$2 	imes 10^{-3} \ A$

Vedclass · Answer

(A) Given:Side of square loop,$l = 10 \ cm = 0.1 \ m$Area of loop,$A = l^2 = (0.1 \ m)^2 = 0.01 \ m^2 = 100 \ cm^2$Resistance,$R = 0.5 \ \Omega$Initial magnetic field,$B_1 = 0.10 \ T$Final magnetic field,$B_2 = 0 \ T$Time interval,$\Delta t = 0.70 \ s$Angle between the area vector (normal to the loop) and the magnetic field vector,$	heta = 45^{\circ}$The magnetic flux $\phi$ is given by $\phi = BA \cos 	heta$.Initial flux,$\phi_1 = B_1 A \cos 45^{\circ} = 0.10 	imes 0.01 	imes \frac{1}{\sqrt{2}} \ Wb$.Final flux,$\phi_2 = 0 \ Wb$.Change in flux,$\Delta \phi = \phi_2 - \phi_1 = -\phi_1 = -\frac{0.001}{\sqrt{2}} \ Wb$.Induced emf,$\varepsilon = -\frac{\Delta \phi}{\Delta t} = -\left( -\frac{0.001}{\sqrt{2} 	imes 0.70} ight) = \frac{0.001}{0.70 	imes 1.414} \approx 1.01 	imes 10^{-3} \ V$.Using the provided approximation in the original problem: $\varepsilon = \frac{0.1 	imes 100 	imes 10^{-4} 	imes \cos 45^{\circ}}{0.70} = \frac{0.001 	imes 0.707}{0.70} \approx 10^{-3} \ V$.Induced current,$I = \frac{\varepsilon}{R} = \frac{10^{-3} \ V}{0.5 \ \Omega} = 2 	imes 10^{-3} \ A$.

Similar Questions

$A$ conducting loop of radius $\frac{10}{\sqrt{\pi}}\,cm$ is placed perpendicular to a uniform magnetic field of $0.5\,T$. The magnetic field is decreased to $zero$ in $0.5\,s$ at a steady rate. The induced emf in the circular loop at $0.25\,s$ is:

The total emf induced in a closely wound coil of $N$ turns in which the magnetic flux linked with the coil is changing at the rate $\frac{d \phi_{B}}{dt}$ is

The induced emf cannot be produced by:

The magnetic flux through a loop of resistance $10 \Omega$ varies according to the relation $\phi = 6t^2 + 7t + 1$,where $\phi$ is in milliweber and time is in seconds. At time $t = 1 \ s$,the induced e.m.f. is:

$A$ square loop of wire,side length $10 \, cm$,is placed at an angle of $45^{\circ}$ with a magnetic field that changes uniformly from $0.1 \, T$ to zero in $0.7 \, s$. The induced current in the loop (its resistance is $1 \, \Omega$) is ..... $mA$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module