Question

A closed coil consists of 500 turns on a rectangular
frame of area 4.0 cm2 and has a resistance of
50 ohms. The coil is kept with its plane
perpendicular to a uniform magnetic field of
0.2 wb/m2, the amount of charge flowing through
the coil if it is turned over (rotated through 180°):-
(1) 1.6 x 10-3 C
(2) 16 x 10-3 C.
(3) 0.16 x 10-3 C
(4) 160 x 10-30​

Answer 1

Answer:

Option (1) $1.6\times 10^{-3}\text{ C}$

Explanation:

No. of turns N = 500

Surface Area S = 4 cm² = 4 × 10⁻⁴ m²

Resistance R = 50 Ohm

Magnetic Field B = 0.2 Wb/m²

Magnetic Density

$\Phi=NBS$

Since the coil is roatated through 180°

Therefore change in the magnetic density

$\Delta \Phi=NBS-(-NBS)=2NBS$

Thus the charge flowing

$\Delta q=\frac{\Delta \Phi}{R}$

or, $\Delta q=\frac{2\times N\times B\times S}{R}$

$\implies \Delta q=\frac{2\times 500\times 0.2\times 4\times 10^{-4}}{50}$

$\implies \Delta q=16\times 10^{-4}\text{ C}$

or, $\Delta q=1.6\times 10^{-3}\text{ C}$

Hope this helps.

Answer 2

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