11. If $\theta_1$ and $\theta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\theta$ is given by: (2017-Delhi)
If planes are at angles $\alpha$ and $90^{\circ} - \alpha$ with the magnetic meridian, then $\tan \theta_1 = \frac{\tan \theta}{\cos \alpha}$ and $\tan \theta_2 = \frac{\tan \theta}{\sin \alpha}$.
Squaring and adding their inverses gives $\cot^2 \theta_1 + \cot^2 \theta_2 = \frac{\cos^2 \alpha + \sin^2 \alpha}{\tan^2 \theta} = \cot^2 \theta$.
12. A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It: (2012 Pre)
At the geomagnetic poles, the horizontal component of Earth's magnetic field is zero ($B_H = 0$). Since the needle is constrained to move only in the horizontal plane, there is no horizontal restoring torque, so it can stay in any direction.
10. At a point A on the earth’s surface the angle of dip, $delta = +25^{circ}$. At a point B on the earth’s surface the angle of dip, $delta = -25^{circ}$. We can interpret that: (2019)
By convention, the angle of dip is taken as positive in the Northern Hemisphere (where magnetic field lines dip downward into the earth) and negative in the Southern Hemisphere (where lines point upward).
11. If $theta_1$ and $theta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $theta$ is given by: (2017-Delhi)
In two mutually perpendicular planes, $tantheta_1 = frac{tantheta}{cosalpha}$ and $tantheta_2 = frac{tantheta}{sinalpha}$. Using the identity $cos^2alpha + sin^2alpha = 1$, we obtain $cot^2theta = cot^2theta_1 + cot^2theta_2$.
1. 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 $\Phi = B A \cos(0^\circ) = 0.5 \times 1 \times 1 = 0.5 Wb$.
2. 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 $t = 4 s$, $e = 10(4) + 3 = 43 V$.
3. 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 $e = |\frac{\Delta \Phi}{\Delta t}| = \frac{2 \times 10^{-3} - 0}{0.1} = 0.02 V$.
8. 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 $|\frac{dr}{dt}| = 1 mm s^{-1} = 10^{-3} m s^{-1}$.
$e = 0.025 \cdot 2\pi(0.02) \cdot (10^{-3}) = \pi \times 10^{-6} V = \pi \mu V$.
9. 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)
10. 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).
$\Phi = \left(\frac{1}{\pi}\right) \cdot (\pi \cdot 0.2^2) \cdot \cos(60^\circ) = 0.04 \cdot 0.5 = 0.02 Wb$.