A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Omega$2\Omega$ then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is ____ N.
A conducting rod AB moves on a U-shaped rail in a perpendicular magnetic field.
A.7.5 times 10^-2$7.5 \times 10^{-2}$
B.5.7 times 10^-3$5.7 \times 10^{-3}$
C.5.7 times 10^-2$5.7 \times 10^{-2}$
D.7.5 times 10^-3$7.5 \times 10^{-3}$
Solution & Explanation
### Related Formula
E = B l v$E = B l v$i = fracER$$i = \frac{E}{R}$$F_B = i l B = fracB^2 l^2 vR$$F_{B} = i l B = \frac{B^2 l^2 v}{R}$$
### Core Logic
To maintain a constant speed, the external force applied must balance the opposing magnetic force generated by the induced current.
F_textext = F_B$$F_{\text{ext}} = F_{B}$$
### Step 1: Calculate External Force
Given values:
B = 0.10text T$B = 0.10\text{ T}$l = 1text m$l = 1\text{ m}$v = 1.5text m/s$v = 1.5\text{ m/s}$R = 2\, Omega$R = 2\, \Omega$F_textext = fracB^2 l^2 vR$$F_{\text{ext}} = \frac{B^2 l^2 v}{R}$$F_textext = frac(0.10)^2 times (1)^2 times 1.52$$F_{\text{ext}} = \frac{(0.10)^2 \times (1)^2 \times 1.5}{2}$$F_textext = frac0.01 times 1.52 = frac0.0152 = 0.0075$$F_{\text{ext}} = \frac{0.01 \times 1.5}{2} = \frac{0.015}{2} = 0.0075$$F_textext = 7.5 times 10^-3text N$$F_{\text{ext}} = 7.5 \times 10^{-3}\text{ N}$$A conducting rod AB moves on a U-shaped rail in a perpendicular magnetic field.
### Pattern Recognition
Standard "sliding rod on rails" problem. The required mechanical force to maintain terminal velocity is always F = fracB^2 L^2 vR$F = \frac{B^2 L^2 v}{R}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Keywords:#metal rod#magnetic field#circuit track
More Electromagnetic Induction Previous-Year Questions — Page 2
Q21jee_main_2025_28_jan_eveningMotional EMF
A conducting \bar moves on two conducting rails as shown in the figure. A constant magnetic field B \exists into the page. The \bar starts to move from the vertex at time t = 0$t = 0$ with a constant velocity. If the induced EMF is E propto t^n$E \propto t^n$ , then value of n$n$ is
A linear conductor \bar moving laterally across V-shaped intersecting conducting guide rails.
Numerical Answer.Answer: 1
Solution
### Related Formula
The motional EMF induced across a moving conductor of instantaneous length ell$\ell$ inside a perpendicular uniform magnetic field is given by:
E = B cdot ell cdot v$$E = B \cdot \ell \cdot v$$
### Core Logic
Let the V-shaped guide rails form an \angle, so that the instantaneous length ell$\ell$ of the conducting \bar grows linearly with its horizontal position distance x$x$ from the vertex [cite: 782, 791]:
ell propto x$\ell \propto x$
Since the \bar moves with a constant velocity v$v$, its displacement position at any time t$t$ is :
x = v cdot t implies ell propto v cdot t$$x = v \cdot t \implies \ell \propto v \cdot t$$
Substituting this time-dependent length into the induced EMF expression :
E = B cdot ell cdot v implies E propto B cdot (v cdot t) cdot v implies E propto t^1$$E = B \cdot \ell \cdot v \implies E \propto B \cdot (v \cdot t) \cdot v \implies E \propto t^1$$
Comparing this to the given relation E propto t^n$E \propto t^n$ gives the exponent[cite: 188, 791]:
n = 1$n = 1$
### Step 1: Geometric Analysis
The expanding circuit loop configuration across time is shown below:
A linear conductor \bar moving laterally across V-shaped intersecting conducting guide rails.
### Pattern Recognition
For \parallel rails, the length ell$\ell$ remains constant, meaning induced EMF is independent of time (E propto t^0$E \propto t^0$). For V-shaped divergent rails, the effective length increases linearly with distance, making the induced EMF directly proportional to time (E propto t^1$E \propto t^1$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Qjee_main_2025_29_jan_morningMutual Inductance
Consider mathrmI_1$\mathrm{I}_1$ and mathrmI_2$\mathrm{I}_2$ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If mathrmL_1 =$\mathrm{L}_1 =$ self inductance of coil 1, mathrmM_12 =$\mathrm{M}_{12} =$ mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be
### Related Formula
phi_1 = L_1 I_1 + M_12 I_2$$\phi_1 = L_1 I_1 + M_{12} I_2$$varepsilon_1 = -fracmathrmdphi_1mathrmdt$$\varepsilon_1 = -\frac{\mathrm{d}\phi_1}{\mathrm{d}t}$$
### Core Logic
The total flux linked with coil 1 is due to its own current I_1$I_1$ and the mutual influence of current I_2$I_2$ in the neighboring coil :
phi_1 = L_1 I_1 + M_12 I_2$$\phi_1 = L_1 I_1 + M_{12} I_2$$
Differentiating with respect to time according to Faraday\'s Law yields :
varepsilon_1 = -L_1 fracmathrmdI_1mathrmdt - M_12 fracmathrmdI_2mathrmdt$$\varepsilon_1 = -L_1 \frac{\mathrm{d}I_1}{\mathrm{d}t} - M_{12} \frac{\mathrm{d}I_2}{\mathrm{d}t}$$
### Pattern Recognition
Total induced emf sums both self-induction and mutual induction effects additively with standard Lenz law negative signs.
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Qjee_main_2025_29_jan_morningAC Generator
A coil of area A and N turns is rotating with angular velocity omega$\omega$ in a uniform magnetic field vecB$\vec{B}$ about an axis perpendicular to vecB$\vec{B}$ . Magnetic flux varphi$\varphi$ and induced emf varepsilon$\varepsilon$ across it, at an instant when vecB$\vec{B}$ is parallel to the plane of coil, are:
### Related Formula
phi = BAN cos(omega t)$$\phi = BAN \cos(\omega t)$$varepsilon = BANomega sin(omega t)$$\varepsilon = BAN\omega \sin(\omega t)$$
### Core Logic
AC Generator explanation diagram for Q12
When the magnetic field vector vecB$\vec{B}$ lines up parallel to the plane of the coil, the norm area vector stands perpendicular to vecB$\vec{B}$, yielding omega t = fracpi2$\omega t = \frac{\pi}{2}$. Thus :
phi = BAN cosleft(fracpi2right) = 0$$\phi = BAN \cos\left(\frac{\pi}{2}\right) = 0$$varepsilon = BANomega sinleft(fracpi2right) = NABomega$$\varepsilon = BAN\omega \sin\left(\frac{\pi}{2}\right) = NAB\omega$$
### Pattern Recognition
Flux is zero when the field lines are parallel to the coil surface, but the rate of change of flux (and thus emf) peaks to its absolute maximum.
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q52jee_main_2024_01_february_morningInduced EMF
A rectangular loop of sides 12mathrm~cm$12\mathrm{~cm}$ and 5mathrm~cm$5\mathrm{~cm}$, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of 5mathrm~cm/s$5\mathrm{~cm/s}$ in the positive x-axis direction, in a space containing a variable magnetic field in the positive z-direction. The field has a gradient of 10^-3mathrm~T/cm$10^{-3}\mathrm{~T/cm}$ along the negative x-direction and it is decreasing with time at the rate of 10^-3mathrm~T/s$10^{-3}\mathrm{~T/s}$. If the resistance of the loop is 6mathrm~mOmega$6\mathrm{~m\Omega}$, the power dissipated by the loop as heat is x times 10^-9mathrm~W$x \times 10^{-9}\mathrm{~W}$. The value of x$x$ is:
Numerical Answer.Answer: 216 to 216
Solution
### Related Formula
Total induced EMF in a moving loop within a time-varying spatial field:
varepsilon_textnet = varepsilon_textmotional + varepsilon_texttime$$\varepsilon_{\text{net}} = \varepsilon_{\text{motional}} + \varepsilon_{\text{time}}$$varepsilon_textmotional = l cdot v cdot Delta B = l cdot v cdot left(fracdBdx cdot Delta xright)$$\varepsilon_{\text{motional}} = l \cdot v \cdot \Delta B = l \cdot v \cdot \left(\frac{dB}{dx} \cdot \Delta x\right)$$varepsilon_texttime = A cdot fracdBdt$$\varepsilon_{\text{time}} = A \cdot \frac{dB}{dt}$$
### Core Logic
Loop dimensions: l = 5mathrm~cm = 0.05mathrm~m$l = 5\mathrm{~cm} = 0.05\mathrm{~m}$, Delta x = 12mathrm~cm = 0.12mathrm~m$\Delta x = 12\mathrm{~cm} = 0.12\mathrm{~m}$.
Velocity v = 5mathrm~cm/s = 0.05mathrm~m/s$v = 5\mathrm{~cm/s} = 0.05\mathrm{~m/s}$.
Spatial gradient fracdBdx = 10^-3mathrm~T/cm = 0.1mathrm~T/m$\frac{dB}{dx} = 10^{-3}\mathrm{~T/cm} = 0.1\mathrm{~T/m}$.
Time decay rate fracdBdt = 10^-3mathrm~T/s$\frac{dB}{dt} = 10^{-3}\mathrm{~T/s}$.
Calculate the motional component across the leading edges:
varepsilon_textmotional = 300 times 10^-7mathrm~V$$\varepsilon_{\text{motional}} = 300 \times 10^{-7}\mathrm{~V}$$
Calculate the time-varying field induction across the loop area:
A = 12 times 5 = 60mathrm~cm^2 = 60 times 10^-4mathrm~m^2$$A = 12 \times 5 = 60\mathrm{~cm}^2 = 60 \times 10^{-4}\mathrm{~m}^2$$varepsilon_texttime = A cdot fracdBdt = 60 times 10^-4 times 10^-3 = 60 times 10^-7mathrm~V$$\varepsilon_{\text{time}} = A \cdot \frac{dB}{dt} = 60 \times 10^{-4} \times 10^{-3} = 60 \times 10^{-7}\mathrm{~V}$$
Both changes induce current in the same direction according to Lenz's law:
varepsilon_textnet = 300 times 10^-7 + 60 times 10^-7 = 360 times 10^-7mathrm~V$$\varepsilon_{\text{net}} = 300 \times 10^{-7} + 60 \times 10^{-7} = 360 \times 10^{-7}\mathrm{~V}$$
### Step 1: Calculate Dissipated Power
Given loop resistance R = 6mathrm~mOmega = 6 times 10^-3mathrm~Omega$R = 6\mathrm{~m\Omega} = 6 \times 10^{-3}\mathrm{~\Omega}$:
P = fracvarepsilon_textnet^2R = frac(360 times 10^-7)^26 times 10^-3 = frac129600 times 10^-146 times 10^-3$$P = \frac{\varepsilon_{\text{net}}^2}{R} = \frac{(360 \times 10^{-7})^2}{6 \times 10^{-3}} = \frac{129600 \times 10^{-14}}{6 \times 10^{-3}}$$P = 21600 times 10^-11 = 216 times 10^-9mathrm~W$$P = 21600 \times 10^{-11} = 216 \times 10^{-9}\mathrm{~W}$$
Therefore, x = 216$x = 216$.
### Pattern Recognition
When a loop moves through a field that changes in both space and time, the total induced EMF is the sum of the motional EMF (vfracpartial Bpartial x$v\frac{\partial B}{\partial x}$) and the transformer EMF (Afracpartial Bpartial t$A\frac{\partial B}{\partial t}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q52jee_main_2024_29_january_eveningMotional Electromotive Force
A horizontal straight wire 5text m$5\text{ m}$ long extending from east to west falling freely at right angle to horizontal component of earth's magnetic field 0.60 times 10^-4text Wb m^-2$0.60 \times 10^{-4}\text{ Wb m}^{-2}$. The instantaneous value of emf induced in the wire when its velocity is 10text ms^-1$10\text{ ms}^{-1}$ is x times 10^-3text V$x \times 10^{-3}\text{ V}$. The value of x$x$ is:
Numerical Answer.Answer: 3 to 3
Solution
### Related Formula
The motional electromotive force (emf) induced in a conductor of length L$L$ moving with velocity v$v$ perpendicular to a magnetic field B$B$ is:
e = B v L$e = B v L$
### Core Logic
Given parameters:
* Length of wire, L = 5text m$L = 5\text{ m}$
* Horizontal magnetic field component, B_H = 0.60 times 10^-4text Wb m^-2$B_H = 0.60 \times 10^{-4}\text{ Wb m}^{-2}$
* Velocity of fall, v = 10text ms^-1$v = 10\text{ ms}^{-1}$
### Step 1: Calculate the Induced EMF
Substitute the parameters directly into the motional emf formula:
e = B_H v L$e = B_H v L$e = (0.60 times 10^-4text Wb m^-2) times (10text ms^-1) times (5text m)$$e = (0.60 \times 10^{-4}\text{ Wb m}^{-2}) \times (10\text{ ms}^{-1}) \times (5\text{ m})$$e = 3 times 10^-3text V$$e = 3 \times 10^{-3}\text{ V}$$
Comparing this to x times 10^-3text V$x \times 10^{-3}\text{ V}$, we find:
x = 3$x = 3$
### Pattern Recognition
Motional EMF is directly the product of field, velocity, and length (e = B v L$e = B v L$) when they are mutually perpendicular. A simple multiplication is all that is required here.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
More Electromagnetic Induction Questions — jee_main_2026_21_jan_morning
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