Explanation
The magnetic field induction at O due to current through portion CD will be zero because the point O lies on the axis of CD when extended towards B.
The magnetic field at the centre o due to current through upper side of semicircular current loop is equal and opposite to that due to lower side of semicircular current loop.
If force on the electron due to electric field is equal and opposite to the force on electron due to magnetic field, then electron will go undeflected.
When current is passed, the spring will contract, breaking the contact with mercury. As contact is broken, magnetic effects become zero. Due to weight the coil will restore its contact with mercury and so on. The coil will have oscillatory behaviour.
Here, I = 4 A ; V = 20 V; so R = V/I = 20/4 = 5A. Since voltmeter is connected in parallel with resistance R, the effective resistance of this combination is 5 ohm only if the resistance R is greater than 5Ω, since total resistance in parallel combination becomes less than individual resistance.
As the currents are in opposite directions, the magnetic field induction due to current in each wire will add up at O. The direction of magnetic field is perpendicular to X – Y plane and is directed inwards i.e. along negative Z – axis.
Force of attraction on the arm SP of loop due to conductor will be stronger than the force of repulsion on arm QR of the loop. Due to which the loop will move towards the conductor.
Both the rings rotate due to magnetic field induction of the other, till both while rotating come along the common plane.
(D) inversely proportional to r
When a charged particle is moving in a magnetic field, its speed does not change but direction of velocity changes.
Apply Right hand thumb rule.
Magnetic field induction at a mid point due to currents through two linear parallel conductors, in the same direction is equal and opposite, hence magnitude of magnetic field induction at that point is zero.
The force on the semicircle part of the loop ADC is equal to the force on the straight part AC of the loop. The length of straight part AC of loop = 2 R. The force on this part= B0 × I × 2R = 2RIB0
The loop will expand due to radially outward force.
Pot. diff. across C and D = 10 V. if x is the resistance of ammeter, then x + R = 10 / 2 = 5 or R = 5 – x < 5.
For non circular trajectory of charged particle in x – y plane, there should be present an electric field as well as magnetic field. If electric field is present along x – axis, then magnetic field should be in z – direction and/or in x –direction but not along y – axis. Therefore, the answer (c) is correct.
Magnetic field lies inside as well as outside the solid current carrying conductor.
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