A coil is placed perpendicular to a magnetic field of 5000 mathrm~T$5000 \mathrm{~T}$. When the field is changed to 3000 mathrm~T$3000 \mathrm{~T}$ in 2mathrms$2\mathrm{s}$, an induced emf of 22 mathrm~V$22 \mathrm{~V}$ is produced in the coil. If the diameter of the coil is 0.02 mathrm~m$0.02 \mathrm{~m}$, then the number of turns in the coil is:
A.7$7$
B.70$70$
C.35$35$
D.140$140$
Solution & Explanation
### Related Formula
varepsilon = N left| fracDeltaphiDelta t right|$\varepsilon = N \left| \frac{\Delta\phi}{\Delta t} \right|$Deltaphi = (Delta B) A costheta$\Delta\phi = (\Delta B) A \cos\theta$
### Core Logic
Given data:
Initial Magnetic Field, B_i = 5000mathrm\,T$B_i = 5000\mathrm{\,T}$
Final Magnetic Field, B_f = 3000mathrm\,T$B_f = 3000\mathrm{\,T}$
Time interval, Delta t = 2mathrm\,s$\Delta t = 2\mathrm{\,s}$
Diameter, d = 0.02mathrm\,m Rightarrow r = 0.01mathrm\,m$d = 0.02\mathrm{\,m} \Rightarrow r = 0.01\mathrm{\,m}$
Induced emf, varepsilon = 22mathrm\,V$\varepsilon = 22\mathrm{\,V}$
Change in magnetic field magnitude |Delta B| = 5000 - 3000 = 2000mathrm\,T$|\Delta B| = 5000 - 3000 = 2000\mathrm{\,T}$.
Area of the coil A = pi r^2 = pi (0.01)^2 = 10^-4pi mathrm\,m^2$A = \pi r^2 = \pi (0.01)^2 = 10^{-4}\pi \mathrm{\,m^2}$.
### Step 2: Equation Evaluation
Deltaphi = |Delta B| A = (2000) pi (0.01)^2 = 0.2pi$\Delta\phi = |\Delta B| A = (2000) \pi (0.01)^2 = 0.2\pi$
Using Faraday's Law:
22 = N left( frac0.2pi2 right)$22 = N \left( \frac{0.2\pi}{2} \right)$22 = N (0.1pi)$22 = N (0.1\pi)$
Taking pi approx 22/7$\pi \approx 22/7$:
22 = N left( 0.1 times frac227 right)$22 = N \left( 0.1 \times \frac{22}{7} \right)$1 = fracN70 Rightarrow N = 70$1 = \frac{N}{70} \Rightarrow N = 70$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
Keywords:#coil is placed perpendicular to a magnetic field#JEE Main 2024 Morning Q44#Electromagnetic Induction JEE Main 2024#Faraday's Law JEE Main 2024
More Electromagnetic Induction Previous-Year Questions — Page 4
Q57jee_main_2024_31_jan_morningMutual Inductance
A small square loop of wire of side ell$\ell$ is placed inside a large square loop of wire of side L$L$ (L = ell^2$L = \ell^2$). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is sqrtx times 10^-7mathrm\ H$\sqrt{x} \times 10^{-7}\mathrm{\ H}$, where x =$x =$
Numerical Answer.Answer: 128 to 128
Solution
### Related Formula
M = fracphi_2i_1$M = \frac{\phi_2}{i_1}$B_textstraight wire segment = fracmu_0 i4pi d (sintheta_1 + sintheta_2)$B_{\text{straight wire segment}} = \frac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$
### Core Logic
Mutual Inductance diagram for Q57 - JEE Main 2024 Morning
Assume a current i$i$ flows through the larger square loop of side L$L$. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side ell$\ell$.
The magnetic field at the center of the large square loop (distance d = L/2$d = L/2$ from each side, angles 45^circ$45^\circ$):
B = 4 times left[ fracmu_0 i4pi (L/2) (sin 45^circ + sin 45^circ) right]$B = 4 \times \left[ \frac{\mu_0 i}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) \right]$B = fracmu_0 ipi (L/2) left( frac2sqrt2 right)$B = \frac{\mu_0 i}{\pi (L/2)} \left( \frac{2}{\sqrt{2}} \right)$B = frac2sqrt2 mu_0 ipi L$B = \frac{2\sqrt{2} \mu_0 i}{\pi L}$
### Step 2: Mutual Inductance Calculation
Flux linkage for the inner loop:
phi = B cdot ell^2$\phi = B \cdot \ell^2$phi = frac2sqrt2 mu_0 ipi L ell^2$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi L} \ell^2$
Given L = ell^2$L = \ell^2$:
phi = frac2sqrt2 mu_0 ipi (ell^2) ell^2 = frac2sqrt2 mu_0 ipi$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi (\ell^2)} \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi}$
Mutual inductance M$M$:
M = fracphii = frac2sqrt2 mu_0pi$M = \frac{\phi}{i} = \frac{2\sqrt{2} \mu_0}{\pi}$
Using mu_0 = 4pi times 10^-7$\mu_0 = 4\pi \times 10^{-7}$:
M = frac2sqrt2 times 4pi times 10^-7pi$M = \frac{2\sqrt{2} \times 4\pi \times 10^{-7}}{\pi}$M = 8sqrt2 times 10^-7mathrm\,H$M = 8\sqrt{2} \times 10^{-7}\mathrm{\,H}$M = sqrt128 times 10^-7mathrm\,H$M = \sqrt{128} \times 10^{-7}\mathrm{\,H}$
Comparing with sqrtx times 10^-7$\sqrt{x} \times 10^{-7}$, we get x = 128$x = 128$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electromagnetic Induction
More Electromagnetic Induction Questions — jee_main_2024_31_jan_morning
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