Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions)
An infinite current carrying conductor is placed along the z axis and a wire looop is kept in the x – y plane. The current in the conductor is increasing with time. Then the
20
Oct
An infinite current carrying conductor is placed along the z axis and a wire looop is kept in the x – y plane. The current in the conductor is increasing with time. Then the An infinite current carrying conductor is placed along the z axis and a wire looop is kept in the x - [...]
A uniform magnetic field exists in a square of side 2a ( as shown in figure ). A square loop of side a enters the field along a diagonal and leave it at a constant speed. Draw the curve between induced emf e and distance along the diagonal , say x0.
20
Oct
A uniform magnetic field exists in a square of side 2a ( as shown in figure ). A square loop of side a enters the field along a diagonal and leave it at a constant speed. Draw the curve between induced emf e and distance along the diagonal , say x0. 0). If the rod [...]
Figures (a) and (b) show a square loop of side a rotating about the given axes in a uniform magnetic field such that the field is perpendicular to the axis in both cases. Angular speed of rotation is both cases being the same.
20
Oct
Figures (a) and (b) show a square loop of side a rotating about the given axes in a uniform magnetic field such that the field is perpendicular to the axis in both cases. Angular speed of rotation is both cases being the same. 0). If the rod moves with a velocity v = v0 find [...]
A conduting wire ABC (as shown) is moving with a constant velocity along horizontal direction. The magnetic field is perpendicular to the wire and directed into the page. AB =BC. Find the range of angle (θ) made by rod AB with horizontal at A so that value of induced emf at A is greater than at C
20
Oct
A conduting wire ABC (as shown) is moving with a constant velocity along horizontal direction. The magnetic field is perpendicular to the wire and directed into the page. AB =BC. Find the range of angle (θ) made by rod AB with horizontal at A so that value of induced emf at A is greater than [...]
Figure shows a conducting frame having battery and a resistance on which a movable conductor of length 0.5 m can slide, The whole arrangement is placed in a uniform magnetic field of B =0.4 T directed perpendicular and into the plane of frame. Initially the circuit is open. When the key is inserted, the conductor begins to move. It is found that a force 0.5 N has to be applied on the conductor to the left to keep it moving at constant speed to the right. Current flowing in the conductor is:
20
Oct
Figure shows a conducting frame having battery and a resistance on which a movable conductor of length 0.5 m can slide, The whole arrangement is placed in a uniform magnetic field of B =0.4 T directed perpendicular and into the plane of frame. Initially the circuit is open. When the key is inserted, the conductor [...]
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Figure shows a conducting frame having battery and a resistance on which a movable conductor of length 0.5 m can slide ,
the conductor begins to move. It is found that a force 0.5 N has to be applied on the conductor to the left to keep it moving at constant speed to the right. Current flowing in the conductor is: ,
The whole arrangement is placed in a uniform magnetic field of B =0.4 T directed perpendicular and into the plane of frame. Initially the circuit is open. When the key is inserted ,
A conducting rod of length 1 is moving on a horizontal smooth surface. Magnetic field in the region is vertically downward and magnitude B0. If centre of mass (COM) of the rod is translating with velocity v0 and rod rotates about COM with angular velocity v0/l, then potential difference between points O and A will be
20
Oct
A conducting rod of length 1 is moving on a horizontal smooth surface. Magnetic field in the region is vertically downward and magnitude B0. If centre of mass (COM) of the rod is translating with velocity v0 and rod rotates about COM with angular velocity v0/l, then potential difference between points O and A will [...]
Figure shows a magnet suspended at the lower end of a spring while its length lies along the axis of a fixed circular coil. THe magnet is made to oscillate . Deflection in the galvanometer is
20
Oct
Figure shows a magnet suspended at the lower end of a spring while its length lies along the axis of a fixed circular coil. THe magnet is made to oscillate . Deflection in the galvanometer is Figure shows a magnet suspended at the lower end of a spring while its length lies along the axis [...]
Figure shows a poerful electromagnet arrangement. A copper ring, which is free to move, is placed on the projecting part of the core as shown. when the key is inserted , the ring
20
Oct
Figure shows a poerful electromagnet arrangement. A copper ring, which is free to move, is placed on the projecting part of the core as shown. when the key is inserted , the ring Figure shows a poerful electromagnet arrangement. A copper ring is placed on the projecting part of the core as shown. when the [...]
An inverted L shaped conductor PRQ is made by joining two perpendicular conducting rods, each of length 1.5 L, at end R. This structure is moving in x-y plane containing variable magnetic field B =−3 xeˆ(k) with a velocity veˆ(i) + v jˆ. If potential of P is Vp and that of Q is VQ, then value of VP − V(Q) at the instant when P is at origin as shown in Fig. Will be
20
Oct
An inverted L shaped conductor PRQ is made by joining two perpendicular conducting rods, each of length 1.5 L, at end R. This structure is moving in x-y plane containing variable magnetic field B =−3 xeˆ(k) with a velocity veˆ(i) + v jˆ. If potential of P is Vp and that of Q is VQ, [...]
Tags:
An inverted L shaped conductor PRQ is made by joining two perpendicular conducting rods ,
at end R. This structure is moving in x-y plane containing variable magnetic field B =−3 xeˆ(k) with a velocity veˆ(i) + v jˆ. If potential of P is Vp and that of Q is VQ ,
each of length 1.5 L ,
then value of VP − V(Q) at the instant when P is at origin as shown in Fig. Will be ,
The magnetic field inn a region is given by B = k B0/L y where L is a fixed length. A conducting rod of length L lies along the Y – axis between the origin and the point (0,L, 0). If the rod moves with a velocity v = v0 , find the emf induced between the ends of the rod.
20
Oct
The magnetic field inn a region is given by B = k B0/L y where L is a fixed length. A conducting rod of length L lies along the Y – axis between the origin and the point (0,L, 0). If the rod moves with a velocity v = v0 , find the emf induced [...]