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Dc pandey electricity and magnetism solutions pdf

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Let charge q flows to C1 at it falls to the free end of the wire. As inward magnetic field is increasing, induced electric field must be anti-clockwise. At point A and B potential is zero not field, hence they may or may not be equilibrium point. Dipole has both negative and positive charges hence work done is not positive. Please inform if this book is available in e-book form for download? As negative charge is at less distance from the line charge, it is attracted towards the line charge. Jul 12, Shivam Bansal added it.

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More than One Correct Options 1. For all charges to be in equilibrium, force experienced by either charge must be zero ie. Hence all the charges must be collinear, charges q1, and q3 must have same sign and q2 must have opposite sign, q2 must have maximum magnitude.

Such on equilibrium is always unstable. Flux through any closed surface depends only on charge inside the surface but electric field at any point on the surface depends on charges inside as well as outside the surface. As net charge on an electric dipole is zero, net flux through the sphere is zero.

But electric field at any point due to a dipole cannot be zero. If two concentric spheres carry equal and opposite charges, Electric field is non-zero only in the region between two sphere and potential is is zero only outside both the spheres. As force on the rod due to electric field is towards right, force on the rod due to hinge must be left. The equilibrium is clearly neutral. If moved along perpendicular bisector, for all identical charges, electrostatic potential energy is maximum at mid point and if moved along the line joining the particles, electrostatic potential energy is minimum at the mid-point.

Charge will flow if there is potential difference between the conductors. It does not depend on amount of charge present.

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Consider the charge distribution shown in figure. Sort-cut Method Entire charge resides on outer surface of conductor and will be divided equally on two outer surfaces.

Let capacitor C1 is initially charged and C2 is uncharged. Hence proved. The arrangement can be considered as the combination of three different capacitors as shown in figure, where. Let C P and CS be the effective capacitance of parallel and series combination respectively. Hence, to increase voltage range to V, 5 capacitors must be connected in series. Hence, a series grouping of 5 such capacitors will have effective capacitance 2 pF and can withstand V.

Given, C1. Charge is shown in mC. Let C1 and C2 be the capacitances of A and B respectively. From Eqs. In this case. Let charge q flows from C1 to C2 and C3. On connecting the plates V becomes zero.

The system can assumed to a parallel combination of two spherical conductors. In the steady state, capacitor behaves as open circuit. P and Q are at same potential, hence capacitor connected between them have no effect on equivalent capacitance.

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Capacitance of all other capacitance is same, e A i. The equivalent circuit is shown in figure. The given arrangement can be considered as the combination of three capacitors as shown in figure.

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Here, plate 1 is connected to plate 5 and plate 3 is connected to plate 6. Reason correctly explains the assertion. Charge on two capacitors will be same only if both the capacitors are initially uncharged. As potential difference across both the capacitors is same, charge will not flow through the switch.

C and R2 are shorted. So more charge flows to C2. When the switch is just closed, capacitors behave like short circuit, no current pass through either 6 W or 5 W resistor.

Charge on capacitor at the given instant. When current in the resistor is 1 A. Hence, no charge will flow from A to B. As potential difference across both the capacitors is same, they are in parallel. Rate of charging decreases as it just charged. Immediately after switch is closed, capacitor behaves like short circuit. Electric field in the gap will remain same. Electric field inside the dielectric slab E V. More Than One Correct Options 1.

Before switch S is closed, charge distribution is shown in figure 1. As magnetic field can exert force on charged particle, it can be accelerated in magnetic field but its speed cannot increases as magnetic force is always perpendicular to the direction of motion of charged particle. As magnetic force provides necessary centripetal force to the particle to describe a circle. Electrons are refocused on x-axis at a distance equal to pitch, i. Consider the disc to be made up of large number of elementary concentric rings.

Consider one such ring of radius x and thickness dx. Charge on this ring. By right hand thumb rule, direction of magnetic field due to conductor A, B, C and D are as shown in figure. Consider one such cylinder of radius r and thickness dr. The component of velocity along the magnetic field i. As initially particle is neutral, charge on two particles must be equal and opposite.

For the electron to hit the target, distance G S must be multiple of pitch, i.

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As the magnetic field is along Y-axis, particle will move in XZ-plane. The path of particle will be a cycloid. In this case, instantaneous centre of curvature of the particle will move along X-axis.

As magnetic force provides centripetal force to the particle, z vz.

DC Pandey Physics Electricity and Magnetism Solution PDF

Consider the disc to be made up of large number of elementary rings. Consider on such ring of radius x and thickness dx. Charge on this ring,. Circuit is same as in Q. Suppose equal and opposite currents are flowing in sides a d and e h, so that three complete current carrying loops are formed,. Net magnetic field at O. Let us consider a point P x, y where magnetic field is zero. Clearly the point must lie either in 1st quadrant or in 3rd quadrant. For magnetic field at the centre of loop to be zero, magnetic field due to straight conductor at centre of loop must be outward, hence I1 must be rightwards.

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At the centre of the loop. Let B1, B2 and B3 be magnetic fields due to complete cylinder, upper and lower cavity respectively. Let us first find magnetic field due a current carrying infinite plate. Objective Questions Level 1 1. Fact 2 pm 2.

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The path will be parabola if force acting on the particle is constant in magnitude as well as in direction. Force acting on a closed current carrying loop is always zero. A current carrying closed loop experiences a force magnetic field. Magnetic field at O due to P,.

Component of velocity of particle along magnetic field, i. Magnetic field due to a conductor of finite length. In case C, magnetic field of conductor and at O is inward while those of and at O is outward, hence net magnetic field at O in this case is zero. Same as question 1 a. At point 1, Magnetic field due to inner conductor is non-zero, but due to outer conductor is zero.

Magnetic field due to straight conductors at O is zero because O lies on axis of both the conductors. Inside a solid cylinder having uniform current density,. For parabolic path, acceleration must be constant and should not be parallel or antiparallel to velocity.

Magnetic force on upper wire must be in upward direction, hence current should be in a direction opposite to that of wire 1. Reason is also correct but does not explain Assertion. Radii of both is different because mass of both is different. Magnetic force cannot change speed of particle as it is always perpendicular to the speed of the particle. Consider the sphere to be made up of large number of hollow, coaxial cylinder of different height and radius. Consider one such cylinder of radius x, height y and thickness.

As solved in question 5 c. Force on portion AC will more compared to that on portion CB. Consider an elementary portion of the wire carrying current I1 of length dx at a distance x from end B. Magnetic field line due to current carrying conductor is shown in figure. Consider an elementary portion of length dy at y - y on the wire.

Force on this portion,. Magnetic force cannot do work on charged particle, hence its energy will remain same, so that q remains same. Again, magnetic force is always along the string, it will never produce a torque hence, T will also remain same. Initially, net force on the particle is zero.

If B1, B 2 and B 3 be magnetic fields at the. As PE U is maximum, equilibrium is unstable. Upward and downward components of force will cancel each other while leftward force is more than rightward force, hence net force is leftwards.

Use right hand thumb rule. For cd to be in equilibrium, force on it must be repulsive while for ab to be in equilibrium, force on it must be attractive. Equilibrium of cd will be stable while that of ab will be unstable. Match the Columns 1. Shashank Holla 15 April at Anonymous 15 April at Tim Railey 12 January at Kushankur Das 25 April at Vineet Sharma 25 April at Shashank Holla 26 April at Aditya Singh 24 July at Unknown 17 September at Anonymous 19 November at Muhammad Saquib Shaikh 20 December at Rishabh Singhal 1 October at