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

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

Question

(A) The area of the square loop is $A = (10 \; cm)^2 = 100 \; cm^2 = 10^{-2} \; m^2$.The angle $	heta$ between the area vector (perpendicular to the

Vedclass · Accepted Answer

$1.0$ mV,$2$ mA

Vedclass · Answer

(A) The area of the square loop is $A = (10 \; cm)^2 = 100 \; cm^2 = 10^{-2} \; m^2$.The angle $	heta$ between the area vector (perpendicular to the east-west plane) and the magnetic field (in the north-east direction) is $45^{\circ}$.The initial magnetic flux is $\phi_i = B A \cos 	heta = 0.10 	imes 10^{-2} 	imes \cos(45^{\circ}) = \frac{10^{-3}}{\sqrt{2}} \; Wb$.The final magnetic flux is $\phi_f = 0 \; Wb$.The change in flux is $\Delta \phi = |\phi_f - \phi_i| = \frac{10^{-3}}{\sqrt{2}} \; Wb$.The magnitude of the induced $emf$ is $\varepsilon = \frac{|\Delta \phi|}{\Delta t} = \frac{10^{-3}}{\sqrt{2} 	imes 0.70} \approx 1.01 	imes 10^{-3} \; V \approx 1.0 \; mV$.The magnitude of the induced current is $I = \frac{\varepsilon}{R} = \frac{1.01 	imes 10^{-3} \; V}{0.5 \; \Omega} \approx 2.02 	imes 10^{-3} \; A \approx 2.0 \; mA$.

Similar Questions

The magnetic field $B$ crossing normally a square metallic plate of area $4\,m^2$ changes with time as shown in the figure. The magnitude of the induced $emf$ in the plate during $t=2\,s$ to $t=4\,s$ is $..........\,mV$.

Assertion : An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current.
Reason : The above statement is in accordance with the conservation of energy.

$A$ coil with an area of $500 \ cm^2$ and $1000$ turns is placed in a magnetic field of $2 \times 10^{-5} \ Wb/m^2$ perpendicular to its plane. If the coil is rotated by $180^o$ in $0.2 \ s$,the induced $emf$ in $milli-volts$ is:

The instantaneous magnetic flux associated with a closed loop of resistance $10 \ \Omega$ is given by $\phi = 2t^2 - 5t + 1$. The magnitude of the induced current at $t = 0.25 \ s$ will be . . . . . . . (in $A$)

$A$ conducting ring of certain resistance is falling towards a current-carrying straight long conductor. The ring and conductor are in the same plane. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module