In series with B.G. of resistance 30 Ω. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction 10-2 T. If it is now turned through an angle of 60° about an axis in its plane. Find the charge induced in the coil. (Area of a coil =10-2 m2)

Question

In series with B.G. of resistance 30 Ω. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction 10-2 T. If it is now turned through an angle of 60° about an axis in its plane. Find the charge induced in the coil. (Area of a coil =10-2 m2) (A) 2×10-5 C (B) 3.2×10-5 C (C) 1×10-5 C (D) 5.5×10-5 C

Tardigrade Admin · Accepted Answer

Given: n =10 turns, R text coil =20 Ω, RG=30 Ω, Total resistance in the circuit =20+30 =50 Ω A=10-2 m 2, B=10-2 T, φ=0°, φ2=60° q=(φ1-φ2/R)=(B n A cos θ1-B n A cos θ2/R) =(B n A( cos 0- cos 60)/R)=(B n A(1-0.5)/R) = (0.5 × 10-2 × 10 × 10-2/50) =(50 × 10-5/50) 1 × 10-5 C

$2 \times 1 0^{- 5} C$

$3.2 \times 1 0^{- 5} C$

$1 \times 1 0^{- 5} C$

$5.5 \times 1 0^{- 5} C$

Q. In series with B.G. of resistance 30Ω. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction 10−2 T. If it is now turned through an angle of 60∘ about an axis in its plane. Find the charge induced in the coil. (Area of a coil =10−2m2)

2×10−5C

3.2×10−5C

1×10−5C

5.5×10−5C

Given : n=10 turns, Rcoil ​=20Ω,RG​=30Ω, Total resistance in the circuit =20+30 =50Ω A=10−2m2,B=10−2T, ϕ=0∘,ϕ2​=60∘ q=Rϕ1​−ϕ2​​=RBnAcosθ1​−BnAcosθ2​​ =RBnA(cos0−cos60)​=RBnA(1−0.5)​ =500.5×10−2×10×10−2​ =5050×10−5​ 1×10−5C

$2 \times 1 0^{- 5} C$

$3.2 \times 1 0^{- 5} C$

$1 \times 1 0^{- 5} C$

$5.5 \times 1 0^{- 5} C$