Two ends of a current carrying wire AB of length l, radius r and resistivity of material ρ are connected to the terminals of a battery of emf E. The wire AB is placed in a uniform magnetic field of induction B→ as shown. Force experienced by the wire is 20 N. It radius of the wire is doubled and length also doubled, force experienced by the wire will be

Two ends of a current carrying wire AB of length l, radius r and resistivity of material $ρ$ are connected to the terminals of a battery of emf E. The wire AB is placed in a uniform magnetic field of induction $\vec{B}$ as shown. Force experienced by the wire is 20 N. It radius of the wire is doubled and length also doubled, force experienced by the wire will be

A
10 N
B
40 N
C
80 N
D
60 N

Solution:

Current through the wire is given by

$\begin{array}{l} i = \frac{E}{R} = \frac{E}{ρ . l / {πr}^{2}} = \frac{E . {πr}^{2}}{ρ . l} \Rightarrow i \propto \frac{r^{2}}{l} \\ F = Bil \\ ∴ F \propto r^{2} \frac{F^{1}}{F} = {(\frac{r^{1}}{r})}^{2} \\ \Rightarrow F^{1} = 4 \times 20 N = 80 N \end{array}$

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