A massless square loop, of wire of resistance 10 Ω, supporting a mass of 1 g, hangs vertically with one of its sides in a uniform magnetic field of 10³ G, directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 1 g? (If sides of the loop = 10 cm, g = 10 m s⁻²) (a) 1 V (b) 1/10 V (c) 10 V (d) 100 V

Answer: (c). Top side (length L = 0.1 m) is in the field B = 10³ G = 0.1 T. Balance: I L B = mg ⇒ I = 10⁻³×100.1×0.1 = 1 A V = IR = 1×10 = 10 V So the answer is (c) 10 V.

A massless square loop, of wire of resistance 10 Ω, supporting a mass… | JEE Main 2023 (30 Jan, Shift 1)

Q32JEE Main 2023 · 30 JAN · SHIFT 1Medium

Force on Current-Carrying Conductors

A massless square loop, of wire of resistance $10\ \Omega$ , supporting a mass of $1\text{ g}$ , hangs vertically with one of its sides in a uniform magnetic field of $10^{3}\text{ G}$ , directed outwards in the shaded region. A dc voltage $\text{V}$ is applied to the loop. For what value of $\text{V}$ , the magnetic force will exactly balance the weight of the supporting mass of $1\text{ g}$ ?

(If sides of the loop $= 10\text{ cm},\ g = 10\text{ m s}^{-2}$ )

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Answer: (c)

Top side (length $L = 0.1\text{ m}$ ) is in the field $B = 10^3\text{ G} = 0.1\text{ T}$ .
Balance: $I L B = mg \Rightarrow I = \frac{10^{-3}\times10}{0.1\times0.1} = 1\text{ A}$
$V = IR = 1\times10 = 10\text{ V}$
So the answer is (c) 10 V.

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