A coil of area 100 cm 2 has 500 turns. Magnetic field of 0.1 weber/metre 2 is perpendicular to the coil. The field is reduced to zero in 0.1 second. The induced e.m.f. in the coil is

Question

A coil of area 100 cm 2 has 500 turns. Magnetic field of 0.1 weber/metre 2 is perpendicular to the coil. The field is reduced to zero in 0.1 second. The induced e.m.f. in the coil is (A) 1 V (B) 5 V (C) 50 V (D) Zero

Tardigrade Admin · Accepted Answer

e=-(N(B2-B1) A cos θ/Δ t) =-(500 ×(0-0.1) × 100 × 10-4 cos 0°/0.1)=5 V

Q. A coil of area $100 c m^{2}$ has $500$ turns. Magnetic field of $0.1$ weber/metre $^{2}$ is perpendicular to the coil. The field is reduced to zero in $0.1$ second. The induced e.m.f. in the coil is

1 V

5 V

50 V

Zero

$e = - \frac{N ( B _{2} - B _{1} ) A c o s θ}{Δ t}$
$= - \frac{500 \times ( 0 - 0.1 ) \times 100 \times 1 0 ^{- 4} c o s 0 ^{\circ}}{0.1} = 5 V$

Q. A coil of area 100cm2 has 500 turns. Magnetic field of 0.1 weber/metre 2 is perpendicular to the coil. The field is reduced to zero in 0.1 second. The induced e.m.f. in the coil is

1 V

5 V

50 V

Zero

e=−ΔtN(B2​−B1​)Acosθ​ =−0.1500×(0−0.1)×100×10−4cos0∘​=5V

Q. A coil of area $100 c m^{2}$ has $500$ turns. Magnetic field of $0.1$ weber/metre $^{2}$ is perpendicular to the coil. The field is reduced to zero in $0.1$ second. The induced e.m.f. in the coil is

$e = - \frac{N ( B _{2} - B _{1} ) A c o s θ}{Δ t}$
$= - \frac{500 \times ( 0 - 0.1 ) \times 100 \times 1 0 ^{- 4} c o s 0 ^{\circ}}{0.1} = 5 V$