$A$ square loop of side $12 \; cm$ with its sides parallel to $X$ and $Y$ axes is moved with a velocity of $8 \; cm \, s^{-1}$ in the positive $x$-direction in an environment containing a magnetic field in the positive $z$-direction. The field is neither uniform in space nor constant in time. It has a gradient of $10^{-3} \; T \, cm^{-1}$ along the negative $x$-direction (that is, it increases by $10^{-3} \; T \, cm^{-1}$ as one moves in the negative $x$-direction), and it is decreasing in time at the rate of $10^{-3} \; T \, s^{-1}$. Determine the direction and magnitude of the induced current in the loop if its resistance is $4.50 \; m\Omega$.
$(2.88 \times 10^{-2} \; A)$ Side of the square loop, $s = 12 \; cm = 0.12 \; m$.
Area of the square loop, $A = 0.12 \times 0.12 = 0.0144 \; m^2$.
Velocity of the loop, $v = 8 \; cm/s = 0.08 \; m/s$.
Gradient of the magnetic field along the negative $x$-direction, $\frac{dB}{dx} = 10^{-3} \; T \, cm^{-1} = 10^{-1} \; T \, m^{-1}$.
Rate of decrease of the magnetic field, $\frac{dB}{dt} = 10^{-3} \; T \, s^{-1}$.
Resistance of the loop, $R = 4.5 \; m\Omega = 4.5 \times 10^{-3} \; \Omega$.
The rate of change of magnetic flux due to the motion of the loop in a non-uniform magnetic field is $\frac{d\phi_1}{dt} = A \times \frac{dB}{dx} \times v = 0.0144 \times 10^{-1} \times 0.08 = 1.152 \times 10^{-4} \; Wb/s$.
The rate of change of flux due to the explicit time variation of the field is $\frac{d\phi_2}{dt} = A \times \frac{dB}{dt} = 0.0144 \times 10^{-3} = 0.144 \times 10^{-4} \; Wb/s$.
The total induced emf is $e = \frac{d\phi_1}{dt} + \frac{d\phi_2}{dt} = 1.152 \times 10^{-4} + 0.144 \times 10^{-4} = 1.296 \times 10^{-4} \; V$.
The induced current is $i = \frac{e}{R} = \frac{1.296 \times 10^{-4}}{4.5 \times 10^{-3}} = 2.88 \times 10^{-2} \; A$.
By Lenz's law, the direction of the current is such that it opposes the change in flux. Since the flux is increasing due to motion and decreasing due to time variation, the net effect results in a clockwise current when viewed from the positive $z$-axis.
Area of the square loop, $A = 0.12 \times 0.12 = 0.0144 \; m^2$.
Velocity of the loop, $v = 8 \; cm/s = 0.08 \; m/s$.
Gradient of the magnetic field along the negative $x$-direction, $\frac{dB}{dx} = 10^{-3} \; T \, cm^{-1} = 10^{-1} \; T \, m^{-1}$.
Rate of decrease of the magnetic field, $\frac{dB}{dt} = 10^{-3} \; T \, s^{-1}$.
Resistance of the loop, $R = 4.5 \; m\Omega = 4.5 \times 10^{-3} \; \Omega$.
The rate of change of magnetic flux due to the motion of the loop in a non-uniform magnetic field is $\frac{d\phi_1}{dt} = A \times \frac{dB}{dx} \times v = 0.0144 \times 10^{-1} \times 0.08 = 1.152 \times 10^{-4} \; Wb/s$.
The rate of change of flux due to the explicit time variation of the field is $\frac{d\phi_2}{dt} = A \times \frac{dB}{dt} = 0.0144 \times 10^{-3} = 0.144 \times 10^{-4} \; Wb/s$.
The total induced emf is $e = \frac{d\phi_1}{dt} + \frac{d\phi_2}{dt} = 1.152 \times 10^{-4} + 0.144 \times 10^{-4} = 1.296 \times 10^{-4} \; V$.
The induced current is $i = \frac{e}{R} = \frac{1.296 \times 10^{-4}}{4.5 \times 10^{-3}} = 2.88 \times 10^{-2} \; A$.
By Lenz's law, the direction of the current is such that it opposes the change in flux. Since the flux is increasing due to motion and decreasing due to time variation, the net effect results in a clockwise current when viewed from the positive $z$-axis.
Explore More
Similar Questions
Two circular coils $P$ and $Q$ are fixed coaxially and carry currents $I_1$ and $I_2$ respectively.
Medium
View Solution$A$ triangular wire frame (each side $= 2 \, m$) is placed in a region of time-variant magnetic field having $dB/dt = \sqrt{3} \, T/s$. The magnetic field is perpendicular to the plane of the triangle. The base of the triangle $AB$ has a resistance $1 \, \Omega$ while the other two sides have resistance $2 \, \Omega$ each. The magnitude of potential difference between the points $A$ and $B$ will be....$V$
$A$ flexible wire bent in the form of a circle is placed in a uniform magnetic field perpendicular to the plane of the coil. The radius of the coil changes as shown in the figure. The graph of induced $emf$ in the coil is represented by
Difficult
View SolutionIn a coil of resistance $10\,\Omega$,the induced current developed by changing magnetic flux through it is shown in the figure as a function of time. The magnitude of change in flux through the coil in weber is:
MediumAIPMT 2012
View Solution$A$ circular coil of $500$ turns of wire has an enclosed area of $0.1\,m^2$ per turn. It is kept perpendicular to a magnetic field of induction $0.2\,T$ and rotated by $180^o$ about a diameter perpendicular to the field in $0.1\,s$. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of $50\,\Omega$?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo