Q82 / 90 · JEE Main 2024 (31 Jan Shift 1) · Physics

A coil is placed perpendicular to a magnetic field of $5000 T$ . When the field is changed to $3000 T$ in $2 s$ , an induced emf of $22 V$ is produced in the coil. If the diameter of the coil is $0.02 m$ , then the number of turns in the coil is:

Solution

Average induced EMF is given by, $ε = N (\frac{∆ ϕ}{t})$

Also, change in flux $∆ ϕ = (∆ B) A$

Given: $B_{i} = 5000 T$ ,

$B_{f} = 3000 T$

$d = 0.02 m \Rightarrow r = \frac{d}{2} = 0.01 m$

Hence,

$∆ ϕ = (∆ B) A$

$= (2000) π {(0.01)}^{2} = 0.2 π$

EMF, $ε = N (\frac{∆ ϕ}{t}) \Rightarrow 22 = N (\frac{0.2 π}{2})$

$\Rightarrow N = 70$

Hence, the correct answer is option: 'B'.