Physics Guru

One edge of a square conducting loop made of uniform wire is mounted on a vertical non-conducting light pole. The light pole is free to rotate about the vertical axis passing through it. A uniform magnetic field $B$ points towards the right side such that loop's plane is parallel to the field as shown in the figure. The current $I_0$ is switched on at $t=0$. Find the instantaneous angular acceleration of the loop at $t=0$. The mass per unit length of the uniform wire is $\mu$ and length of one edge is $l$. Solution Force $F$ on the right edge $F=I_0lB$. This force will create torque which will rotate the loop. Torque $\tau =l.F=I_0 l^2B$ We have $\tau = I\alpha$ $I=\mu l.l^2+\mu l.\frac {l^2}{3}+\mu l.\frac {l^2}{3}$ $\therefore I=\frac {5}{3} \mu l^3 $ Now, $\tau =l.F=I_0 l^2B = I \alpha = \frac {5}{3} \mu l^3 \alpha$ $\therefore \alpha = \frac {3I_0B}{5\mu l}$

During an excavation in the southern part of India, a bundle of old leaves bearing inscriptions was discovered. During carbon dating, a sample of the leaves was found to have a C-14 activity of 14.40 disintegrations per gram per minute. A fresh, identical sample of the leaves has a C-14 activity of 15.12 disintegrations per gram per minute. Determine the age of the old leaves. The half-life of C-14 is 5730 years. Solution We have, $t = 3.32{t_{1/2}}{\log _{10}}\frac{{{{[A]}_0}}}{{[A]}}$ $\Rightarrow t = 3.32 \times 5730 \times {\log _{10}}\frac{{15.12}}{{14.40}}$ $ \Rightarrow t = 19023.6 \times {\log _{10}}1.05$ $ \Rightarrow t \approx 19023.6 \times 0.02119 \approx 403$ years

A small projectile is launched with velocity $u$ in $xz$ vertical plane in a gravity-free large hall from origin $(0, 0, 0)$ at an angle of $60^\circ$ from $x-axis$. At what position on $x-axis$ should another projectile be launched at the same instant with identical speed $u$ in the $xz$ plane at an angle of $120^\circ$ from $x-axis$ so that the two projectiles collide after time $t$? (ignore air resistance) Answer:  As can be seen in the figure, we have an equilateral triangle.  The distance between the two, $OA = OP = AP = ut$.  So, the position of A $\equiv (ut, 0, 0)$.

A small decorative 50 mW LED lamp is connected to a 3.0 V DC button cell. The circuit is left open, with a 2.0 cm gap between two exposed copper terminals inside a terracotta diya. The LED is mounted at the top of the diya, while the button cell is hidden at the bottom. Person X proposes pouring ordinary tap water into the diya so that the water bridges the gap between the two terminals. He says that distilled water is generally considered safe but is of no use here. Person Y says tap water causes a short circuit. So, pouring it would burn the circuit. Who is correct? Answer Usually, it is considered that water and electricity do not play well together, and water is a no-no as far as electricity is concerned. This perception has a lot to do with high voltages, including domestic supply voltages, and the dangers posed by water during the rainy season or in bathrooms. In this problem, however, we are dealing with a low-voltage, small-current situation. Tap water is not a short circuit in...

In the circuit shown, the battery is connected across AB. Will the drift velocity of electrons be maximum in the smallest resistance $R$ as there would be least resistance for the electrons to move in $R$? (all resistors are made of wires with identical cross sectional area and identical material) Answer The given circuit is actually balanced Wheatstone bridge as $\frac {4R}{2R} = \frac {8R}{4R}$ which means no current flows through the resistance $R$. So, even though the resistance $R$ is smallest, the drift velocity is not going to be maximum in $R$. In fact the drift velocity would be zero in $R$.  In the top we would have $2R+4R=6R$ and in the bottom we would have $4R+8R=12R$. $6R$ and $12R$ would be in parallel, so the higher current would flow through $6R$. So, the current would be maximum in $2R$ and $4R$ both.  Now, $I=nAev_d$ So, drift velocity is same for resistive wires having identical current, cross-sectional area and identical material. Hence, the drift velocity ...

Q: Why does the axis of rotation of a spinning top tilt further away from the vertical after some time? (Neglect air friction. Assume real-life situation otherwise.) A: Friction at the pivot point drains the top's rotational kinetic energy, causing its spin speed to drop over time. The spin angular momentum $L=Iω$ therefore decreases as the top slows down. Reduced angular momentum means less resistance to a change in the direction of the axis. In a real situation, there are always small moments when the axis is not absolutely vertical. In such instances, gravity exerts a torque that tends to tip the top over. The top responds to this torque by precessing—the axis rotates around the vertical. As the spin speed decreases, the rate of precession increases and the motion becomes increasingly pronounced. As the top tilts farther, the gravitational torque also increases because the perpendicular distance from the pivot to the line of action of gravity increases. With less angular momentu...

Consider a charged parallel plate capacitor without dielectric. An electric field exists directed from the positive plate toward the negative plate. When a dielectric slab is brought in, its molecular response depends on its nature: If the molecules are non-polar, the external field displaces their electron clouds toward the positive plate, inducing dipole moments. If the molecules are already polar, the existing dipoles rotate to align with the field, with the negative end of each dipole orienting toward the positive plate and the positive end toward the negative plate. After this reorientation, the dielectric can be visualized as a chain of aligned dipoles. A layer of bound negative charge accumulates adjacent to the positive plate, and a layer of bound positive charge accumulates adjacent to the negative plate. These bound charges produce an internal electric field within the dielectric that opposes the original field. The superposition of the two fields results in a reduced net ele...