Question
A conducting rod of 1 meter length and $1 \mathrm{~kg}$ mass is suspended by two vertical wires through its ends. An external magnetic field of 2 Tesla is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is [take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$(a) $0.5 \mathrm{Amp}$(b) $15 \mathrm{Amp}$(c) $5 \mathrm{Amp}$(d) $1.5 \mathrm{Amp}$
Step 1
Step 1: The magnetic force on the rod is given by $F = BIL$, where $B$ is the magnetic field, $I$ is the current, and $L$ is the length of the rod. Show more…
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A conducting rod of $1 \mathrm{~m}$ length and $1 \mathrm{~kg}$ mass is suspended by two vertical wires through the ends. An external magnetic fields of $2 \mathrm{~T}$ is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is (Take $\left.g=10 \mathrm{~ms}^{-2}\right) \quad[$ Kerala CET 2007] (a) $0.5 \mathrm{~A}$ (b) $15 \mathrm{~A}$ (c) $5 \mathrm{~A}$ (d) $1.5 \mathrm{~A}$ (e) $15 \mathrm{~A}$
Magnetic Effect of Current
Round 2
A straight wire of mass $200 \mathrm{gm}$ and length $1.5$ meter carries a current of 2 Amp. It is suspended in mid-air by a uniform horizontal magnetic field B. [take $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]$. The $\mathrm{B}$ is (a) $\overline{(2 / 3) \text { tes } 1 a}$ (b) $(3 / 2)$ tesla (c) $(20 / 3)$ tesla (d) (3/20) tesla
A straight wire of mass $200 \mathrm{gm}$ and length $1.5 \mathrm{~m}$ carries a current of $2 \mathrm{~A}$. It is suspended in mid-air by a uniform horizontal magnetic field $B$. The magnitude of $B$ (in tesla) is $$ \left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right) $$ (A) 2 (B) $1.5$ (C) $0.55$ (D) $0.66$
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