Param Himalaya - परम हिमालय

सतत आवेश वितरण (Continuous Charge Distribution) परिभाषा जब आवेश किसी वस्तु पर असंख्य छोटे-छोटे भागों में फैला हो (न कि एक या कुछ बिंदुओं पर), तो उसे सतत आवेश वितरण कहते हैं। > जैसे — किसी तार, चादर या ठोस गोले पर फैला आवेश सतत आवेश वितरण के प्रकार 1. रेखीय आवेश वितरण (Linear Charge Distribution) जब आवेश किसी रेखा या तार पर फैला हो। रेखीय आवेश घनत्व (λ): $\lambda = \frac{dq}{dl}$ $\Rightarrow dq = \lambda \cdot dl$ - मात्रक (Unit):C/m (कूलॉम प्रति मीटर) - उदाहरण: आवेशित तार, छड़ > कुल आवेश: $Q = \int \lambda \, dl$ 2. पृष्ठीय आवेश वितरण (Surface Charge Distribution) जब आवेश किसी पृष्ठ (surface) पर फैला हो। पृष्ठीय आवेश घनत्व (σ): $\sigma = \frac{dq}{dA}$ $\Rightarrow dq = \sigma \cdot dA$ - मात्रक (Unit): C/m² (कूलॉम प्रति वर्ग मीटर) - उदाहरण: आवेशित चादर, गोले की सतह > कुल आवेश: $Q = \int \sigma \, dA$ 3. आयतनिक आवेश वितरण (Volume Charge Distribution) जब आवेश किसी आयतन (volume) में फैला हो। आयतनिक आवेश घनत्व (ρ): $\rho = \frac{dq}{dV}$ $\Rightarrow dq = \rho \cdot dV$ - मात्रक...

ALTERNATING VOLTAGE APPLIED TO A CAPACITOR Alternating source connected to a capacitor of capacitance C. Such a circuit is known as purely capacitive circuit. The capacitor is periodically charged and discharged when alternating voltage is applied to it. The alternating voltage applied across the capacitor is given by   $V = V_0 \sin \omega t \quad 1$ Let q be the charge on the capacitor at any instant. Then, potential difference across the capacitor,   $V_C = \frac{q}{C}$ But   $V_C = V \quad \text{or} \quad \frac{q}{C} = V = V_0 \sin \omega t$ $\therefore q = V_0 C \sin \omega t$ Now,   $I = \frac{dq}{dt} = \frac{d}{dt}(V_0 C \sin \omega t)$ $I= V_0 C (\cos \omega t) \omega$ $I= \frac{V_0}{(1 / C \omega)} \cos \omega t$ Since   $\cos \omega t = \sin \left(\omega t + \frac{\pi}{2}\right)$ $I= \frac{V_0}{(1 / C \omega)} \sin \left(\omega t + \frac{\pi}{2}\right)$ $\therefore I = I_0 \sin \left(\omega t + \frac{\pi}{2}\right) \quad ...(2)...

Alternating Voltage Applied to an Inductor :  An alternating source is shown connected to an ideal inductor of inductance (L). Such a circuit is known as purely inductive circuit. The alternating voltage across the inductor is given by   $V = V_0 \sin \omega t \quad \text{…(i)}$ The induced e.m.f. across the inductor =  $- L \frac{dI}{dt}$ which opposes the growth of current in the circuit. As there is no potential drop across the circuit, so,   $V + \left(-L \frac{dI}{dt}\right) = 0$ $L \frac{dI}{dt} = V$ $\frac{dI}{dt} = \frac{V}{L}$ Using eqn. (i), we get,   $\frac{dI}{dt} = \frac{V_0}{L} \sin \omega t$ $dI = \frac{V_0}{L} \sin \omega t \, dt \quad \text{…(ii)}$ Integrating both sides, we get,   $\int dI = \int \frac{V_0}{L} \sin \omega t \, dt = \frac{V_0}{L} \int \sin \omega t \, dt$ or   $I = \frac{V_0}{L} \left(-\frac{\cos \omega t}{\omega}\right) = \frac{V_0}{L \omega} (-\cos \omega t)$ Since   $(-\cos \omega ...

Define magnetic flux. State its S.I. unit and write its dimensional formula  Defination of Magnetic Flux : It is defined as the number of magnetic field lines passing through a surface. It is denoted by . Consider a surface S in a uniform magnetic field $\vec{B}$. Let the surface is made up of small elements each of area vector $\vec{dS}$ . If the magnetic field makes an angle with the area vector $\vec{dS}=(ds) \hat{n}$ then the magnetic flux linked with the small element is given by $d\phi_B = \vec{B} \cdot d\vec{S} \quad ...(1)$ The magnetic flux through the surface S is equal to the sum of magnetic flux linked with all elements each of area vector $\vec{dS}$ of the surface. That is, $\phi_B = \sum \vec{B} \cdot d\vec{S} \quad ...(2)$ If the elements are very small in area and extremely large in number, then the magnetic flux linked with the closed surface is written as $\phi_B = \oint_S \vec{B} \cdot d\vec{S} \quad ...(3)$ S.I Unit of Magnetic Flux :  S.I. unit of magne...

Derivation : Potential energy of a magnetic dipole placed in magnetic field. Potential energy of magnetic dipole:   Work done to rotate a magnetic dipole in a uniform magnetic field is stored as potential energy of the magnetic dipole. If a magnetic dipole of magnetic dipole moment \( \vec{m} \) is placed at an angle \( \theta \) with respect to uniform magnetic field of strength \( \vec{B} \), then torque experienced by it is given by: \[\tau = mB \sin \theta\] If the dipole rotates through an angle \( d\theta \), then work done on it is: \[dW = \tau d\theta = mB \sin \theta d\theta\] Total work done to rotate the dipole from \( \theta_1 \) to \( \theta_2 \) position is: $W = \int_{\theta_1}^{\theta_2} mB \sin \theta d\theta$ $W= mB \int_{\theta_1}^{\theta_2} \sin \theta d\theta$ $W= mB [-\cos \theta]_{\theta_1}^{\theta_2}$ $W= -mB [\cos \theta_2 - \cos \theta_1]$ This work done is stored as the potential energy \( U \) of the magnetic dipole: $\therefore U = W = -mB (\cos \...

MAGNETIC DIPOLE (BAR MAGNET) IN A UNIFORM MAGNETIC FIELD :  Derive an expression for torque acting on a magnetic dipole placed in magnetic field. Consider a magnetic dipole (a bar magnet) placed in a uniform magnetic field \( \vec{B} \) such that the angle between the direction of magnetic dipole moment \( (\vec{m}) \) and the direction of magnetic field \( (\vec{B}) \) is \( \theta \)  Let:   - Magnetic Length of magnet = \( 2l \)   - Pole Strength of each pole = \( q_m \)   - Force acting on North pole = \( q_m B \) along the direction of \( \vec{B} \)   - Force acting on South pole = \( q_m B \) opposite to the direction of \( \vec{B} \) These two equal and opposite forces constitute a couple which tends to rotate the magnet in the direction of \( \vec{B} \). Thus, the bar magnet experiences a torque and tends to rotate. However, net force acting on the magnetic dipole is zero and hence magnetic dipole does not have the translational ...

Magnetic Field Intensity at a Point on Equatorial Line of Bar Magnet : Let (P) be a point on the equatorial line of a bar magnet such that its distance from centre (O) is (r). Magnetic field intensity at ( P ) due to N-pole of the bar magnet: $\vec{B_1} = \frac{\mu_0}{4\pi} \cdot \frac{q_m}{(\sqrt{r^2 + l^2})^2} \quad \text{along PC}$ $= \frac{\mu_0}{4\pi} \cdot \frac{qm}{(r^2 + l^2)} \quad \text{along PC} \quad ...(i)$ Magnetic field intensity at ( P ) due to S-pole of the bar magnet: $\vec{B_2} = \frac{\mu_0}{4\pi} \cdot \frac{q_m}{(r^2 + l^2)} \quad \text{along PS} \quad ...(ii)$ Resultant Magnetic Field on Equatorial Line $ ( \vec{B_1}$ ) and $( \vec{B_2} )$ are inclined at an angle $(2\theta)$.   Therefore, the resultant of these two field intensities is: $B_e = \sqrt{B1^2 + B2^2 + 2B1B_2 \cos 2\theta}$ Since $( |\vec{B1}| = |\vec{B2}|$: $B_e = \sqrt{2B1^2 + 2B_1^2 \cos 2\theta}$ $= \sqrt{2B_1^2 (1 + \cos 2\theta)}$ $(\because 1 + \cos 2\theta = 2 \cos^2 \theta)$ $= \sqrt...