Question 61:
easyInduced emf produced in a coil of area \(A\) rotating with angular velocity \(\omega\) in a magnetic field \(B\) is equal to \[\frac{\sqrt{3}BA\omega}{2}\], when the angle between the area vector of the coil and direction of magnetic field is
The induced emf is given by \(e = BA\omega \sin\theta\). Given \(e = \frac{\sqrt{3}}{2}BA\omega\), we have \(\sin\theta = \frac{\sqrt{3}}{2}\), which yields \(\theta = \frac{\pi}{3}\).
Question 62:
easy27. Among which the magnetic susceptibility does not depend on the temperature: (2001)
Diamagnetism is an intrinsic property arising from the orbital motion of electrons and does not involve the alignment of permanent atomic dipoles against thermal agitation. Hence, its susceptibility is independent of temperature.
Question 63:
easy1. A square loop of side $1 m$ and resistance $1 \Omega$ is placed in a magnetic field of $0.5 T$. If the plane of loop of perpendicular to the direction of a magnetic field, the magnetic flux through the loop is: (2022)
Area of the square loop is $A = 1 \times 1 = 1 m^2$.
Since the plane of the loop is perpendicular to the magnetic field, the normal to the loop is parallel to the field, so $\theta = 0^\circ$.
Magnetic flux is $\Phi = B A \cos(0^\circ) = 0.5 \times 1 \times 1 = 0.5 Wb$.
Question 64:
easy2. The magnetic flux linked with a coil (in Wb) is given by the equation $\phi = 5t^2 + 3t + 16$. The magnitude of induced emf in the coil at the fourth second will be: (2020-Covid)
Magnitude of induced emf is $e = |\frac{d\phi}{dt}|$.
Differentiating flux with respect to time: $\frac{d\phi}{dt} = 10t + 3$.
At $t = 4 s$, $e = 10(4) + 3 = 43 V$.
Question 65:
easy3. A 800 turn coil of effective area $0.05 m^2$ is kept perpendicular to a magnetic field of $5 \times 10^{-5} T$. When the plane of the coil is rotated by $90^\circ$ around any of its coplanar axis in $0.1 s$, the emf induced in the coil will be: (2019)
Initial flux is $\Phi_1 = N B A \cos(0^\circ) = 800 \times (5 \times 10^{-5}) \times 0.05 = 2 \times 10^{-3} Wb$.
After rotating by $90^\circ$, $\Phi_2 = N B A \cos(90^\circ) = 0$.
Magnitude of induced emf is $e = |\frac{\Delta \Phi}{\Delta t}| = \frac{2 \times 10^{-3} - 0}{0.1} = 0.02 V$.
Question 66:
easy8. A conducting circular loop is placed in a uniform magnetic field, $B = 0.025 T$ with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of $1 mm s^{-1}$. The induced emf when the radius is $2 cm$ is: (2010 Pre)
Flux is $\Phi = B \cdot A = B \cdot \pi r^2$. Magnitude of induced emf is $e = |\frac{d\Phi}{dt}| = B \cdot 2\pi r |\frac{dr}{dt}|$.
Given $B = 0.025 T$, $r = 2 cm = 0.02 m$, and $|\frac{dr}{dt}| = 1 mm s^{-1} = 10^{-3} m s^{-1}$.
$e = 0.025 \cdot 2\pi(0.02) \cdot (10^{-3}) = \pi \times 10^{-6} V = \pi \mu V$.
Question 67:
easy9. A conducting circular loop is placed in a uniform magnetic field $0.04 T$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $2 mm/s$. The induced emf in the loop when the radius is $2 cm$ is: (2009)
Induced emf $e = |\frac{d}{dt}(B \pi r^2)| = B \pi \cdot 2r |\frac{dr}{dt}|$.
Substitute $B = 0.04 T$, $r = 0.02 m$, and $|\frac{dr}{dt}| = 2 \times 10^{-3} m/s$.
$e = 0.04 \times \pi \times 2(0.02) \times 2 \times 10^{-3} = 3.2\pi \times 10^{-6} V = 3.2\pi \mu V$.
Question 68:
easy10. A circular disc of radius $0.2 meter$ is placed in a uniform magnetic field of induction $\frac{1}{\pi} (\frac{wb}{m^2})$ in such a way that its axis makes an angle of $60^\circ$ with the magnetic field. The magnetic flux linked with the disc is: (2008)
Magnetic flux is given by $\Phi = BA \cos\theta$.
Here, $\theta = 60^\circ$ (angle between the axis, which is the normal to the area, and the magnetic field).
$\Phi = \left(\frac{1}{\pi}\right) \cdot (\pi \cdot 0.2^2) \cdot \cos(60^\circ) = 0.04 \cdot 0.5 = 0.02 Wb$.
Question 69:
easy12. The magnetic flux through a circuit of resistance $R$ changes by an amount $\Delta \phi$ in a time $\Delta t$. Then the total quantity of electric charge $Q$ that passes any point in the circuit during the time $\Delta t$ is represented by: (2004)
The induced emf is $e = \frac{\Delta \phi}{\Delta t}$.
The induced current is $I = \frac{e}{R} = \frac{\Delta \phi}{R \Delta t}$.
Total charge $Q = I \Delta t = \frac{\Delta \phi}{R \Delta t} \cdot \Delta t = \frac{\Delta \phi}{R}$.
Question 70:
easy13. For a coil having $L = 2 mH$, current flow through it is $I = t^2 e^{-t}$ then the time at which emf become zero: (2001)
Induced emf is $e = -L \frac{dI}{dt}$. For $e = 0$, we must have $\frac{dI}{dt} = 0$.
Differentiating $I$: $\frac{dI}{dt} = \frac{d}{dt}(t^2 e^{-t}) = 2t e^{-t} - t^2 e^{-t} = t e^{-t}(2 - t)$.
Setting $\frac{dI}{dt} = 0$ gives $t = 2 s$ (for $t > 0$).