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A galvanometer of resistance 3663 Ω gives full scale deflection for a certain current Ig. The value of the resistance of the shunt which when joined to the galvanometer coil will result in 1/34 of the total current passing through the galvanometer is
Let the shunt resistance be Rs and galvanometer resistance be Rg = 3663Ω Given: only 1/34 of total current I passes through the galvanometer. Using current division rule: Ig/Is = Rs/Rg ' Where Ig/I = 1/34' ⇒ Is/I = 33/34 So: 1/33 = Rs/3663 ⇒ Rs = 3663/33 = 111 Ω ≈100Ω Answer: (A) 100 Ω.
Let the shunt resistance be
Rs and galvanometer resistance be
Rg = 3663Ω
Given: only 1/34 of total current I passes through the galvanometer.
Using current division rule:
Ig/Is = Rs/Rg ‘ Where Ig/I = 1/34’ ⇒ Is/I = 33/34
So:
1/33 = Rs/3663 ⇒ Rs = 3663/33 = 111 Ω ≈100Ω
Answer: (A) 100 Ω.
See lessA galvanometer with resistance 1000 Ω gives full scale deflection at 0.1 mA. What value of resistance should be added to 1000 Ω to increase its current range 10 A.
To convert the galvanometer into an ammeter reading up to 10 A, a small shunt resistance must be added in parallel, allowing most current to bypass the galvanometer. The galvanometer allows only 0.1 mA. Using the formula Rs = Ig.Rg/I-Ig, we get Rs ≈ 0.01Ω. Correct answer: (B) 0.01 Ω in parallel. ForRead more
To convert the galvanometer into an ammeter reading up to 10 A, a small shunt resistance must be added in parallel, allowing most current to bypass the galvanometer. The galvanometer allows only 0.1 mA. Using the formula Rs = Ig.Rg/I-Ig, we get Rs ≈ 0.01Ω.
Correct answer: (B) 0.01 Ω in parallel.
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Two unlike magnetic poles of strength 10 A-m each are held in air at a distance of O- IO m from each other. What is the magnetic force of attraction between them ?
The magnetic force F between two magnetic poles is given by: F = μ0 ⋅m1 ⋅m2/4πr² Here, μ0 = 4π × 10⁻⁷ N/A² ,m = m = 10A\cdotpm, r = 0.1m. Substituting the values: F = 4π × 10⁻⁷ . 10 . 10/4π. (0.1)² = 10⁻⁵/0.01 = 1 × 10⁻³N Answer: (D) 1 × 10⁻³ N So, the magnetic force of attraction is 1 × 10⁻³ N.
The magnetic force F between two magnetic poles is given by:
F = μ0 ⋅m1 ⋅m2/4πr²
Here,
μ0 = 4π × 10⁻⁷ N/A² ,m = m = 10A\cdotpm,
r = 0.1m.
Substituting the values:
F = 4π × 10⁻⁷ . 10 . 10/4π. (0.1)² = 10⁻⁵/0.01 = 1 × 10⁻³N
Answer: (D) 1 × 10⁻³ N
So, the magnetic force of attraction is
See less1 × 10⁻³ N.
A bar magnet AB with magnetic moment M is cut into two equal parts perpendicular to its axis. One part is kept over the other so that end B is exactly over A. What will be the magnetic moment of the combination so formed ?
When a bar magnet of magnetic moment M is cut into two equal parts perpendicular to its length, each part gets half the length but retains the same pole strength. Hence, the magnetic moment of each piece becomes M/2. Placing one over the other with opposite poles touching (B over A), their magneticRead more
When a bar magnet of magnetic moment
M is cut into two equal parts perpendicular to its length, each part gets half the length but retains the same pole strength. Hence, the magnetic moment of each piece becomes M/2. Placing one over the other with opposite poles touching (B over A), their magnetic moments oppose and cancel out. Answer: (A) Zero.
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The value of the magnetic susceptibility for a super conductor is :
The magnetic susceptibility (χ) of a superconductor is highly negative, typically χ =−1. This is because superconductors exhibit perfect diamagnetism—they completely expel magnetic fields from their interior (Meissner effect). As a result, they oppose applied magnetic fields strongly, indicating a lRead more
The magnetic susceptibility (χ) of a superconductor is highly negative, typically
χ =−1. This is because superconductors exhibit perfect diamagnetism—they completely expel magnetic fields from their interior (Meissner effect). As a result, they oppose applied magnetic fields strongly, indicating a large negative susceptibility.
Answer: (D) Highly negative.
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