Q113 / 125 · BITSAT 2014 · Physics
A coil 10 turns and a resistance of $20 \Omega$ is connected in series with B.G of resistance $30 \Omega$. The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction $10^{-2} \mathrm{~T}$. If it is now turned through an angle of $60^{\circ}$ about an axis in its plane. Find the charge induced in the coil. (Area of a coil $=10^{-2} \mathrm{~m}^{2}$ )
- A$2 \times 10^{-5} \mathrm{C}$
- B$3.2 \times 10^{-5} \mathrm{C}$
- Correct answerC$1 \times 10^{-5} \mathrm{C}$
- D$5.5 \times 10^{-5} \mathrm{C}$
Solution
Given: $\mathrm{n}=10$ turns, $\mathrm{R}_{\text {coil }}=20 \Omega, \mathrm{R}_{\mathrm{G}}=30 \Omega$, Total resistance in the circuit $=20+30$
$=50 \Omega$
$\mathrm{A}=10^{-2} \mathrm{~m}^{2}, \mathrm{~B}=10^{-2} \mathrm{~T}, \phi_{1}=0^{\circ}, \phi_{2}=60^{\circ}$
$\mathrm{q}=\frac{\phi_{1}-\phi_{2}}{\mathrm{R}}=\frac{\mathrm{Bn} \mathrm{A} \cos \theta_{1}-\mathrm{Bn} \mathrm{A} \cos \theta_{2}}{\mathrm{R}}$
$=\frac{\operatorname{Bn} \mathrm{A}(\cos 0-\cos 60)}{\mathrm{R}}=\frac{\mathrm{BnA}(1-0.5)}{\mathrm{R}}$
$=\frac{0.5 \times 10^{-2} \times 10 \times 10^{-2}}{50}=\frac{50 \times 10^{-5}}{50}$
$=1 \times 10^{-5} \mathrm{C} \quad$ (Charge induced in a coil)
Hence, the correct answer is option: 'C'.