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 3
Q42jee_main_2024_27_jan_morningFaraday's Law
A rectangular loop of length 2.5text m$2.5\text{ m}$ and width 2text m$2\text{ m}$ is placed at 60^circ$60^{\circ}$ to a magnetic field of 4text T$4\text{ T}$. The loop is removed from the field in 10text sec$10\text{ sec}$. The average emf induced in the loop during this time is:
A.-2text V$-2\text{ V}$
B.+2text V$+2\text{ V}$
C.+1text V$+1\text{ V}$
D.-1text V$-1\text{ V}$
Solution
### Related Formula
textemf = -fracDeltaphiDelta t$$\text{emf} = -\frac{\Delta\phi}{\Delta t}$$
### Core Logic
Initial magnetic flux is given by phi_i = B A costheta$\phi_i = B A \cos\theta$, where A = 2.5 times 2 = 5text m^2$A = 2.5 \times 2 = 5\text{ m}^2$, B = 4text T$B = 4\text{ T}$, and theta = 60^circ$\theta = 60^{\circ}$ (angle aligned with the axis mapping context rules in the problem source text).
phi_i = 4 times 5 times cos(60^circ) = 20 times 0.5 = 10text Wb$$\phi_i = 4 \times 5 \times \cos(60^{\circ}) = 20 \times 0.5 = 10\text{ Wb}$$
Final flux after removal phi_f = 0$\phi_f = 0$.
### Step 1: Compute Induced EMF
textemf = -fracphi_f - phi_iDelta t = -frac0 - 1010 = +1text V$$\text{emf} = -\frac{\phi_f - \phi_i}{\Delta t} = -\frac{0 - 10}{10} = +1\text{ V}$$
### Pattern Recognition
Removal fields yield positive flux variants under canonical sign configurations due to the absolute reduction profile mapped by Lenz/Faraday relationships.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q53jee_main_2024_27_jan_morningMutual Induction
Two coils have mutual inductance 0.002text H$0.002\text{ H}$. The current changes in the first coil according to the relation i = i_0sinomega t$i = i_{0}\sin\omega t$, where i_0 = 5text A$i_{0} = 5\text{ A}$ and \omega = 50pitext rad/s$50\pi\text{ rad/s}$. The maximum value of emf in the second coil is fracpialphatext V$\frac{\pi}{\alpha}\text{ V}$. The value of alpha$\alpha$ is ______.
Numerical Answer.Answer: 2 to 2
Solution
### Related Formula
textemf = -M fracdidt$$\text{emf} = -M \frac{di}{dt}$$
### Core Logic
Differentiating the current expression with respect to time:
fracdidt = fracddt(i_0 sinomega t) = i_0 omega cosomega t$$\frac{di}{dt} = \frac{d}{dt}(i_0 \sin\omega t) = i_0 \omega \cos\omega t$$
Hence, the expression for induced emf is:
textemf = -M i_0 omega cosomega t$$\text{emf} = -M i_0 \omega \cos\omega t$$
### Step 1: Isolate Maximum value
The peak magnitude occurs when cosomega t = 1$\cos\omega t = 1$:
textemf_textmax = M i_0 omega$$\text{emf}_{\text{max}} = M i_0 \omega$$
Substitute given numerical values:
textemf_textmax = 0.002 times 5 times 50pi = 0.5pi = fracpi2text V$$\text{emf}_{\text{max}} = 0.002 \times 5 \times 50\pi = 0.5\pi = \frac{\pi}{2}\text{ V}$$
### Step 2: Match with target variable
Comparing fracpi2$\frac{\pi}{2}$ with fracpialpha$\frac{\pi}{\alpha}$ gives:
alpha = 2$\alpha = 2$
### Pattern Recognition
Harmonic driving functions induce derivative cosine arrays whose peak amplitudes scale strictly as the product of primary parameters: M cdot i_0 cdot omega$M \cdot i_0 \cdot \omega$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Q53jee_main_2024_29_jan_morningFaraday's and Lenz's Law
A square loop of side 10 mathrm~cm$10 \mathrm{~cm}$ and resistance 0.7 \, Omega$0.7 \, \Omega$ is placed vertically in east-west plane. A uniform magnetic field of 0.20 mathrm~T$0.20 \mathrm{~T}$ is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 mathrm~s$1 \mathrm{~s}$ at a steady rate. Then, magnitude of induced emf is sqrtmathrmx times 10^-3 mathrm~V$\sqrt{\mathrm{x}} \times 10^{-3} \mathrm{~V}$. The value of x is
Numerical Answer.Answer: 2 to 2
Solution
### Related Formula
According to Faraday's Law of Electromagnetic Induction, the magnitude of induced EMF (e$e$) is:
e = fracDelta phiDelta t$$e = \frac{\Delta \phi}{\Delta t}$$
where magnetic flux phi$\phi$ is defined via dot product:
phi = vecB cdot vecA = B A cos theta$$\phi = \vec{B} \cdot \vec{A} = B A \cos \theta$$
### Core Logic
Let the East-West plane lie vertical along the x-z plane. The normal vector to the loop points along the North direction (along hatj$\hat{j}$):
vecA = (0.1 mathrm~m)^2 hatj = 0.01 hatj mathrm~m^2$$\vec{A} = (0.1 \mathrm{~m})^2 \hat{j} = 0.01 \hat{j} \mathrm{~m^2}$$
The uniform magnetic field points North-East, meaning it is at an angle of 45^circ$45^\circ$ to the North vector direction:
vecB = B cos 45^circ hati + B sin 45^circ hatj = frac0.2sqrt2hati + frac0.2sqrt2hatj$$\vec{B} = B \cos 45^\circ \hat{i} + B \sin 45^\circ \hat{j} = \frac{0.2}{\sqrt{2}}\hat{i} + \frac{0.2}{\sqrt{2}}\hat{j}$$Vector reference showing orientation of loop area and North-East magnetic field vectors for Q53
### Step 1: Calculate Initial Flux
$phi_i = vecB cdot vecA = left( frac0.2sqrt2 right) times 0.01 = frac2 times 10^-3sqrt2 = sqrt2 times 10^-3 mathrm~Wb$$$\phi_i = \vec{B} \cdot \vec{A} = \left( \frac{0.2}{\sqrt{2}} \right) \times 0.01 = \frac{2 \times 10^{-3}}{\sqrt{2}} = \sqrt{2} \times 10^{-3} \mathrm{~Wb}$$
### Step 2: Find Induced EMF
Since the field goes steadily to zero in $
### Step 2: Find Induced EMF
Since the field goes steadily to zero in $\Delta t = 1 \mathrm{~s}, the final flux value $, the final flux value $\phi_f = 0:
$:
$e = frac|0 - phi_i|1 = sqrt2 times 10^-3 mathrm~V$e = \frac{|0 - \phi_i|}{1} = \sqrt{2} \times 10^{-3} \mathrm{~V}$
### Step 3: Extract x
Comparing this with the target expression $
### Step 3: Extract x
Comparing this with the target expression $\sqrt{x} \times 10^{-3} \mathrm{~V}:
$:
$x = 2
### Pattern Recognition
Be careful when identifying angles between directional plane descriptors. An East-West plane has a normal axis directed North-South. A North-East field makes a clean $
### Pattern Recognition
Be careful when identifying angles between directional plane descriptors. An East-West plane has a normal axis directed North-South. A North-East field makes a clean $45^\circ$ angle relative to this structural normal line.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
A power transmission line feeds input power at 2.3 mathrm~kV$2.3 \mathrm{~kV}$ to a step down transformer with its primary winding having 3000$3000$ turns. The output power is delivered at 230 mathrm~V$230 \mathrm{~V}$ by the transformer. The current in the primary of the transformer is 5 mathrmA$5 \mathrm{A}$ and its efficiency is 90\%$90\%$. The winding of transformer is made of copper. The output current of transformer is ________ mathrmA$\mathrm{A}$.
Numerical Answer.Answer: 45 to 45
Solution
### Related Formula
eta = fracP_textoutP_textin$$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$$P_textin = V_p I_p$$P_{\text{in}} = V_p I_p$$P_textout = V_s I_s$$P_{\text{out}} = V_s I_s$$
### Core Logic
The efficiency eta$\eta$ of a transformer is the ratio of output power to input power. We can use this to find the secondary (output) current.
### Step 1: Calculate Input Power
Given:
V_p = 2.3 mathrm~kV = 2300 mathrm~V$V_p = 2.3 \mathrm{~kV} = 2300 \mathrm{~V}$I_p = 5 mathrm~A$I_p = 5 \mathrm{~A}$P_textin = 2300 times 5 mathrm~W$$P_{\text{in}} = 2300 \times 5 \mathrm{~W}$$
### Step 2: Calculate Output Power and Current
Efficiency eta = 90\% = 0.9$\eta = 90\% = 0.9$P_textout = eta times P_textin = 0.9 times 2300 times 5$$P_{\text{out}} = \eta \times P_{\text{in}} = 0.9 \times 2300 \times 5$$
Since P_textout = V_s I_s$P_{\text{out}} = V_s I_s$ and V_s = 230 mathrm~V$V_s = 230 \mathrm{~V}$:
V_s I_s = 230 times I_s = 0.9 times 2300 times 5$$V_s I_s = 230 \times I_s = 0.9 \times 2300 \times 5$$I_s = frac0.9 times 2300 times 5230 = 0.9 times 10 times 5$$I_s = \frac{0.9 \times 2300 \times 5}{230} = 0.9 \times 10 \times 5$$I_s = 9 times 5 = 45 mathrm~A$$I_s = 9 \times 5 = 45 \mathrm{~A}$$
### Pattern Recognition
Transformer equations are direct: eta V_p I_p = V_s I_s$\eta V_p I_p = V_s I_s$. The number of turns (3000$3000$) is distractor data not needed unless calculating the secondary turns.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Alternating Current
Q60jee_main_2024_30_jan_morningMotional EMF in Rotating Blades
A ceiling fan having 3 blades of length 80 mathrm~cm$80 \mathrm{~cm}$ each is rotating with an angular velocity of 1200 mathrm~rpm$1200 \mathrm{~rpm}$. The magnetic field of earth in that region is 0.5 mathrm~G$0.5 \mathrm{~G}$ and angle of dip is 30^circ$30^{\circ}$. The emf induced across the blades is N pi times 10^-5 mathrm~V$N \pi \times 10^{-5} \mathrm{~V}$. The value of N$N$ is _______.
Numerical Answer.Answer: 32 to 32
Solution
### Related Formula
varepsilon = frac12 B_v omega ell^2$$\varepsilon = \frac{1}{2} B_v \omega \ell^2$$B_v = B sin delta$$B_v = B \sin \delta$$
### Core Logic
For a horizontal fan rotating in the Earth's magnetic field, the blades cut only the vertical component of the magnetic field (B_v$B_v$). The number of blades is a distractor, as they are all in parallel, meaning the EMF developed across one blade is the same as the EMF across the whole fan setup.
### Step 1: Calculate Field and Omega
Vertical component of magnetic field:
B_v = B sin(30^circ) = (0.5 times 10^-4 mathrm~T) times frac12 = 0.25 times 10^-4 mathrm~T = frac14 times 10^-4 mathrm~T$$B_v = B \sin(30^\circ) = (0.5 \times 10^{-4} \mathrm{~T}) \times \frac{1}{2} = 0.25 \times 10^{-4} \mathrm{~T} = \frac{1}{4} \times 10^{-4} \mathrm{~T}$$
Angular velocity in rad/s:
omega = 2pi f = 2pi left(frac120060right) = 40pi mathrm~rad/s$$\omega = 2\pi f = 2\pi \left(\frac{1200}{60}\right) = 40\pi \mathrm{~rad/s}$$
### Step 2: Calculate Induced EMF
Length of blade ell = 80 mathrm~cm = 0.8 mathrm~m$\ell = 80 \mathrm{~cm} = 0.8 \mathrm{~m}$.
varepsilon = frac12 B_v omega ell^2$$\varepsilon = \frac{1}{2} B_v \omega \ell^2$$varepsilon = frac12 left(frac14 times 10^-4right) (40pi) (0.8)^2$$\varepsilon = \frac{1}{2} \left(\frac{1}{4} \times 10^{-4}\right) (40\pi) (0.8)^2$$varepsilon = frac18 times 10^-4 times 40pi times 0.64$$\varepsilon = \frac{1}{8} \times 10^{-4} \times 40\pi \times 0.64$$varepsilon = 5pi times 10^-4 times 0.64$$\varepsilon = 5\pi \times 10^{-4} \times 0.64$$varepsilon = 3.2pi times 10^-4 mathrm~V$$\varepsilon = 3.2\pi \times 10^{-4} \mathrm{~V}$$varepsilon = 32pi times 10^-5 mathrm~V$$\varepsilon = 32\pi \times 10^{-5} \mathrm{~V}$$
### Step 3: Extract N
Given varepsilon = Npi times 10^-5 mathrm~V$\varepsilon = N\pi \times 10^{-5} \mathrm{~V}$, we can directly see that N = 32$N = 32$.
### Pattern Recognition
Motional EMF for rotating rods acts like a battery. Multiple identical blades radiating from the center to the rim behave like multiple identical batteries in parallel; the total voltage does not stack. Isolate B_v$B_v$ for horizontal spinners.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Class 12 Physics: Magnetism and Matter
More Electromagnetic Induction Questions — jee_main_2026_21_jan_morning
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