Let the length of each wire be L. For the square, side = L/4; area = (L/4)² = L²/16. For the circle, radius = L/2π; area = π(L/2π)² = L² /4π. Ratio of magnetic moments = L²/4π/L²/16 = 16/4π = ...
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Voltage sensitivity is given by ls/R’ where I s is current sensitivity. If current sensitivity increases by 20% and resistance by 25%, voltage sensitivity changes by 120/125 = 0.96, i.e., decreases by 4%. The answer is (D) Decrease by ...
To remove the tension in the supporting strings, the magnetic force (BIl) should balance the gravitational force (mg). Therefore, equating BIl = mg, the current required is I = mg/Bl. Hence, the Correct answer is (c) mg/lB.
A current-carrying wire kept in a uniform magnetic field will experience a maximum force when it is perpendicular to the magnetic field. This is because the magnetic force is given by F = Bill sin θ and sin 90° = ...
If an ammeter is to be used as a voltmeter, a high resistance must be connected in series with it. This combination limits the current through the ammeter while allowing it to measure the potential difference accurately.