JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.

**Q2.**An electron is accelerated from rest through a potential difference V. This electron experiences a force F in a uniform magnetic field. On increasing the potential difference to V', the force experienced by the electron in the same magnetic field becomes 2F. Then, the ratio (V^'/V) is equal to

**Q3.**A charged particle moving along +ve x-direction with a velocity v enters a region where there is a uniform magnetic field B_0 (-k ̂ ), from x=0 to x=d. The particle gets deflected at an angle Î¸ from its initial path. The specific charge of the particle is

Solution

**Q4.**A long cylindrical wire of radius ‘a’ carries a current i distributed uniformly over its cross section. If the magnetic fields at distances r<a and R<a from the axis have equal magnitude, then

Solution

**Q5.**A Q5.charged particle moves with velocity v =ai ̂+dj ̂ in a magnetic field B =Ai ̂+Dj ̂. The force acting on the particle has magnitude F. Then,

**Q6.**A very thin metallic wires placed along X-and Y-axes carry equal currents as shown in figure AB and CD are lines at 45° with the axes. The magnetic fields will be zero on the line

**Q7.**A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius b. The spiral lies in the X-Y plane and a steady current I flows through the wire. The Z-component of the magnetic field at the centre of the spiral is

**Q8.**An electron is projected at an angle Î¸ with a uniform magnetic field. If the pitch of the helical path is equal to its radius, then the angle of projection is

**Q9.**An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to this current at the point M is H_1. Now, another infinitely long straight conductor QS is connected to Q so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H_2. The ratio H_1/H_2 is given by

**Q10.**A small block of mass m, having charge q, is placed on a frictionless inclined plane making an angle Î¸ with the horizontal. There exists a uniform magnetic field B parallel to the inclined plane but perpendicular to the length of spring. If m is slightly pulled on the incline in downward direction, the time period of oscillation will be (assume that the block does not leave contact with the plane)

Solution