In a region of uniform magnetic induction B=10-2 tesla, a circular coil of radius 30 cm and resistance π2 ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm, the amplitude of the alternating current induced in the coil is

Question

In a region of uniform magnetic induction B=10-2 tesla, a circular coil of radius 30 cm and resistance π2 ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm, the amplitude of the alternating current induced in the coil is (A) 4 π2 mA (B) 30 mA (C) 6 mA (D) 200 mA

Tardigrade Admin · Accepted Answer

Amplitude of the current, i0=(e0/R)=(ω N B A/R)=(2 π v N B(π r2)/R) i0=(2 π × (200/60) × 1 × 10-2 × π(0.3)2/π2) =6 × 10-3 A =6 mA

$4 π^{2} m A$

$30 m A$

$6 m A$

$200 m A$

Amplitude of the current,
$i_{0} = \frac{e _{0}}{R} = \frac{ω NB A}{R} = \frac{2 π v NB ( π r ^{2} )}{R}$
$i_{0} = \frac{2 π \times \frac{200}{60} \times 1 \times 1 0 ^{- 2} \times π ( 0.3 ) ^{2}}{π ^{2}}$
$= 6 \times 1 0^{- 3} A = 6 m A$

4π2mA

30mA

6mA

200mA

Amplitude of the current, i0​=Re0​​=RωNBA​=R2πvNB(πr2)​ i0​=π22π×60200​×1×10−2×π(0.3)2​ =6×10−3A=6mA

$4 π^{2} m A$

$30 m A$

$6 m A$

$200 m A$

Amplitude of the current,
$i_{0} = \frac{e _{0}}{R} = \frac{ω NB A}{R} = \frac{2 π v NB ( π r ^{2} )}{R}$
$i_{0} = \frac{2 π \times \frac{200}{60} \times 1 \times 1 0 ^{- 2} \times π ( 0.3 ) ^{2}}{π ^{2}}$
$= 6 \times 1 0^{- 3} A = 6 m A$