The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of 24 \, Omega$24 \, \Omega$ is applied. The resistance of galvanometer coil will be:
A.12 \, Omega$12 \, \Omega$
B.96 \, Omega$96 \, \Omega$
C.48 \, Omega$48 \, \Omega$
D.100 \, Omega$100 \, \Omega$
Solution & Explanation
### Related Formula
For a shunted galvanometer, the potential difference across the galvanometer is equal to the potential difference across the shunt resistor:
I_g G = (I - I_g) S$I_g G = (I - I_g) S$
### Core Logic
Let the current sensitivity per division be x$x$.
* Initially, without the shunt, the total current I$I$ gives a full scale deflection of 25 divisions:
I = 25x$I = 25x$Schematic diagrams representing galvanometer before and after shunting for Q44
* When the shunt (S = 24 \, Omega$S = 24 \, \Omega$) is connected in parallel, the current passing through the galvanometer branch (I_g$I_g$) corresponds to a deflection of 5 divisions:
I_g = 5x$I_g = 5x$Schematic diagrams representing galvanometer before and after shunting for Q44
### Step 1: Find Current through the Shunt
The current remaining for the shunt branch is:
I - I_g = 25x - 5x = 20x$I - I_g = 25x - 5x = 20x$
### Step 2: Equate Potential Drops
Applying the parallel branch condition:
5x cdot G = 20x cdot 24$5x \cdot G = 20x \cdot 24$
Dividing both sides by 5x$5x$:
G = 4 times 24 = 96 \, Omega$G = 4 \times 24 = 96 \, \Omega$
Therefore, the resistance of the galvanometer coil is 96 \, Omega$96 \, \Omega$.
### Pattern Recognition
The deflection is directly proportional to the branch current. A drop from 25 divisions to 5 divisions means only frac525 = frac15$\frac{5}{25} = \frac{1}{5}$ of the total current flows through the galvanometer, leaving frac45$\frac{4}{5}$ to go through the shunt. Hence, the galvanometer resistance must be 4 times the shunt resistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Keywords:#deflection in moving coil galvanometer falls#JEE Main 2024 Morning Q44#Magnetism JEE Main 2024#Galvanometer and its Conversions
More Moving Charges and Magnetism Previous-Year Questions — Page 3
Q21jee_main_2025_24_jan_eveningSolenoid
A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75ns. The number of turns per metre in the solenoid is ____. The figure details a thick solenoid cylinder with a internal cross section displaying a charge tracking loop.
[Take mass of electron m_e = 9 times 10^-31$m_e = 9 \times 10^{-31}$ kg, charge of electron |q_e| = 1.6 times 10^-19$|q_e| = 1.6 \times 10^{-19}$ C, mu_0 = 4pi times 10^-7 fracNA^2, 1 text ns = 10^-9 text s$\mu_0 = 4\pi \times 10^{-7} \frac{N}{A^2}, 1 \text{ ns} = 10^{-9} \text{ s}$]
Numerical Answer.Answer: 250 to 250
Solution
### Related Formula
Time period of a revolving charge in a magnetic field:
T = frac2pi mqB$T = \frac{2\pi m}{qB}$
Magnetic field inside a long solenoid:
B = mu_0 n I$B = \mu_0 n I$
### Core Logic
Combining the expressions to isolate n$n$ (turns per meter):
T = frac2pi mq(mu_0 n I)$T = \frac{2\pi m}{q(\mu_0 n I)}$
Substituting the given constants:
75 times 10^-9 = frac2pi times 9 times 10^-311.6 times 10^-19 times 4pi times 10^-7 times n times 1.5$75 \times 10^{-9} = \frac{2\pi \times 9 \times 10^{-31}}{1.6 \times 10^{-19} \times 4\pi \times 10^{-7} \times n \times 1.5}$
Simplifying terms:
75 times 10^-9 = frac18pi times 10^-319.6pi times 10^-26 times n = frac1.875 times 10^-5n$75 \times 10^{-9} = \frac{18\pi \times 10^{-31}}{9.6\pi \times 10^{-26} \times n} = \frac{1.875 \times 10^{-5}}{n}$n = frac1.875 times 10^-575 times 10^-9 = 250$n = \frac{1.875 \times 10^{-5}}{75 \times 10^{-9}} = 250$
### Pattern Recognition
The circular motion time period depends exclusively on the field magnitude B$B$, completely independent of the orbit's velocity or radius.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Q24jee_main_2025_24_jan_morningMagnetic Field due to a Current Element
A current of 5A exists in a square loop of side frac1sqrt2text m$\frac{1}{\sqrt{2}}\text{ m}$ Then the magnitude of the magnetic field B at the centre of the square loop will be ptimes10^-6text T$p\times10^{-6}\text{ T}$ where, value of p is [Take mu_0=4pitimes10^-7text T mA^-1$\mu_{0}=4\pi\times10^{-7}\text{ T mA}^{-1}$].
Numerical Answer.Answer: 8 to 8
Solution
### Related Formula
The magnetic field B_1$B_{1}$ produced by a straight wire segment carrying current I$I$ at a perpendicular distance d$d$ is given by the Biot-Savart relation:
B_1 = fracmu_0I4pi d(sintheta_1 + sintheta_2)$B_{1} = \frac{\mu_{0}I}{4\pi d}(\sin\theta_{1} + \sin\theta_{2})$
### Core Logic
As shown in the square geometric layout Magnetic Field due to a Current Element diagram for Q24 - JEE Main 2025 Morning, the perpendicular distance from any side to the central origin point is exactly half the total side length :
d = fraca2 = frac12sqrt2text m$d = \frac{a}{2} = \frac{1}{2\sqrt{2}}\text{ m}$
Connecting the ends of a side to the center forms internal angles of theta_1 = theta_2 = 45^circ$\theta_{1} = \theta_{2} = 45^{\circ}$.
### Step 1: Summing the Contributions
Calculate the magnetic field contribution from a single side :
B_1 = frac10^-7 times 5frac12sqrt2 left(sin 45^circ + sin 45^circ
ight) = 10^-7 times 10sqrt2 times left(frac2sqrt2
ight) = 2 times 10^-6text T$B_{1} = \frac{10^{-7} \times 5}{\frac{1}{2\sqrt{2}}} \left(\sin 45^{\circ} + \sin 45^{\circ}
ight) = 10^{-7} \times 10\sqrt{2} \times \left(\frac{2}{\sqrt{2}}
ight) = 2 \times 10^{-6}\text{ T}$
Since the current flows in the same rotational direction along all four sides, their individual magnetic fields add constructively at the center :
B_textnet = 4 times B_1 = 4 times (2 times 10^-6text T) = 8 times 10^-6text T$B_{\text{net}} = 4 \times B_{1} = 4 \times (2 \times 10^{-6}\text{ T}) = 8 \times 10^{-6}\text{ T}$
Comparing this with p times 10^-6text T$p \times 10^{-6}\text{ T}$ , we get:
p = 8$p = 8$
### Pattern Recognition
The magnetic field at the center of any square loop simplifies to the standard formula: B = frac2sqrt2mu_0Ipi a$B = \frac{2\sqrt{2}\mu_{0}I}{\pi a}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Qjee_main_2025_29_jan_morningAmpere\'s Circuital Law
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire\'s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
A galvanometer has a resistance of 50mathrm~Omega$50\mathrm{~\Omega}$ and it allows maximum current of 5mathrm~mA$5\mathrm{~mA}$. It can be converted into voltmeter to measure upto 100mathrm~V$100\mathrm{~V}$ by connecting in series a resistor of resistance:
A.5975mathrm~Omega$5975\mathrm{~\Omega}$
B.20050mathrm~Omega$20050\mathrm{~\Omega}$
C.19950mathrm~Omega$19950\mathrm{~\Omega}$
D.19500mathrm~Omega$19500\mathrm{~\Omega}$
Solution
### Related Formula
Voltmeter series conversion formula:
V = I_g(R_g + R)$V = I_g(R_g + R)$R = fracVI_g - R_g$R = \frac{V}{I_g} - R_g$
### Core Logic
Given data:
R_g = 50mathrm~Omega$R_g = 50\mathrm{~\Omega}$, I_g = 5mathrm~mA = 5 times 10^-3mathrm~A$I_g = 5\mathrm{~mA} = 5 \times 10^{-3}\mathrm{~A}$, target voltage range V = 100mathrm~V$V = 100\mathrm{~V}$.
Substitute values:
R = frac1005 times 10^-3 - 50$R = \frac{100}{5 \times 10^{-3}} - 50$
### Step 1: Complete Arithmetic Evaluation
R = 20000 - 50 = 19950mathrm~Omega$R = 20000 - 50 = 19950\mathrm{~\Omega}$
### Pattern Recognition
Voltmeter resistance is always high because it is connected in parallel to circuits to prevent current drawing leaks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
Class 12 Physics: Current Electricity
Q51jee_main_2024_01_february_morningMagnetic Field due to a Current Element
A regular polygon of 6 sides is formed by bending a wire of length 4pi$4pi$ meter. If an electric current of 4pisqrt3mathrm~A$4pisqrt{3}\mathrm{~A}$ is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be x times 10^-7mathrm~T$x \times 10^{-7}\mathrm{~T}$. The value of x$x$ is:
Numerical Answer.Answer: 72 to 72
Solution
### Related Formula
Magnetic field due to a straight wire segment of length 2L$2L$ at distance r$r$:
B_1 = fracmu_0 I4pi r(sintheta_1 + sintheta_2)$B_1 = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)$
Total field for a regular hexagon (n=6$n=6$):
B = 6 times B_1$B = 6 \times B_1$
### Core Logic
Total perimeter = 6 cdot a = 4pi implies textside length a = frac4pi6 = frac2pi3mathrm~m$= 6 \cdot a = 4\pi \implies \text{side length } a = \frac{4\pi}{6} = \frac{2\pi}{3}\mathrm{~m}$.
For a regular hexagon segment, the interior angles relative to the normal vector are theta_1 = theta_2 = 30^circ$\theta_1 = \theta_2 = 30^\circ$.
The normal distance r$r$ from the center to a side is:
r = fraca2 cot(30^circ) = frac4pi2 times 6 times sqrt3 = fracsqrt3pi3 = fracpisqrt3mathrm~m$r = \frac{a}{2} \cot(30^\circ) = \frac{4\pi}{2 \times 6} \times \sqrt{3} = \frac{\sqrt{3}\pi}{3} = \frac{\pi}{\sqrt{3}}\mathrm{~m}$
### Step 1: Calculate Total Magnetic Field
Substitute r$r$ and I = 4pisqrt3mathrm~A$I = 4\pi\sqrt{3}\mathrm{~A}$ into the hexagon configuration:
B = 6 times left[ fracmu_0 I4pi r (sin(30^circ) + sin(30^circ)) right]$B = 6 \times \left[ \frac{\mu_0 I}{4\pi r} (\sin(30^\circ) + \sin(30^\circ)) \right]$B = 6 times left[ frac10^-7 times 4pisqrt3left(fracsqrt3pi3right) times (0.5 + 0.5) right]$B = 6 \times \left[ \frac{10^{-7} \times 4\pi\sqrt{3}}{\left(\frac{\sqrt{3}\pi}{3}\right)} \times (0.5 + 0.5) \right]$B = 6 times left[ frac10^-7 times 4pisqrt3 times 3sqrt3pi times 1 right]$B = 6 \times \left[ \frac{10^{-7} \times 4\pi\sqrt{3} \times 3}{\sqrt{3}\pi} \times 1 \right]$B = 6 times [ 4 times 3 times 10^-7 ] = 6 times 12 times 10^-7 = 72 times 10^-7mathrm~T$B = 6 \times [ 4 \times 3 \times 10^{-7} ] = 6 \times 12 \times 10^{-7} = 72 \times 10^{-7}\mathrm{~T}$
Thus, x = 72$x = 72$.
### Pattern Recognition
For regular polygons, the normal distance r$r$ to the side is always r = fraca2cot(fracpin)$r = \frac{a}{2}\cot(\frac{\pi}{n})$. The contribution from all n$n$ symmetric segments adds up constructively.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Moving Charges and Magnetism
More Moving Charges and Magnetism Questions — jee_main_2024_29_jan_morning
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