Force on a Current Element
Trending Questions
Q.
Charge moving with uniform velocity produce?
Q. A straight wire carrying a current i1 runs along the axis of a circular current i2. Then the force of interaction between the two current carrying conductors is
- ∞
- Zero
- μ04π2i1i2r
- 2i1i2rNm
Q. A square loop ABCD, carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be
- 2μ0IiL3π
- 2μ0Ii3π
- μ0Ii2π
- μ0IiL2π
Q. A conductor lies along the z-axis at–1.5≤z≤1.5 m and carries a fixed current of 10.0 A in –^k direction (see figure). For a field →B=3.0×10−4e0.2x^j T, find the power required to move the conductor at a constant speed to x=2.0 m, y=0 m in 5×10–3 s. Assume parallel motion along the x-axis. (Assume e0.4=1.5)
- 14.85 W
- 29.7 W
- 1.57 W
- 4.5 W
Q. Figure shows a square loop. 20 cm on each side in the x-y plane with its centre at the origin. The loop carries a current of 7 A.
Above it at y=0, z=12 cm is an infinitely long wire parallel to the x-axis carrying a current of 10 A. The net force on the loop is ———×10–5 N.
Above it at y=0, z=12 cm is an infinitely long wire parallel to the x-axis carrying a current of 10 A. The net force on the loop is ———×10–5 N.
Q. A particle with charge q and mass m is projected with kinetic energy K into the region between two plates of a uniform magnetic field B as shown below.
If the particle is to miss collision with the opposite plate, the maximum value of B is
If the particle is to miss collision with the opposite plate, the maximum value of B is
- √2Kqmd
- √2Kdqm
- √2Kmqd
- √2Kqmd
Q.
A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field , such that is perpendicular to the plane of the loop. The magnetic force acting on the loop is?
[BIT 1992; MP PET 1994; IIT 1983;MP PMT 1999; AMU (Engg.) 2000]
Zero
Q. In the figure, AB is a long straight wire carrying a current of 20A and CDFG is a rectangular loop of size 20 cm x 9cm carrying a current of 10A. The edge CG is parallel to AB, at a distance of 1cm from it. The force exerted on the loop by the magnetic field of the wire is
- 3.6×10−4N attraction
- 3.6×10−4N epulsion
- 7.2×10−4N attraction
- 7.2×10−4N repulsion
Q. Two parallel wires aligned with y- axis carrying equal currents in opposite directions are placed at x=+a and x=−a in the x-y plane. The net magnetic field at the origin, O is B1 and at a point P (2a, 0, 0) is B2. Then the ratio B1:B2 is:
- 3:1
- 1:2
- 1:3
- 2:1
Q. An elastic circular wire of length l carries a current I. It is placed in a uniform magnetic field →B (Out of paper) such that its plane is perpendicular to the direction of →B. The wire will experience
- No force
- A stretching force
- A compressive force
- A torque
Q. An equilateral triangular coil of 10 turns carrying 1A current is placed in uniform magnetic field of 4T as shown in figure. If side of triangle is 1m then torque acting on coil is
Q. A conductor lies along the z-axis at–1.5≤z≤1.5 m and carries a fixed current of 10.0 A in –^k direction (see figure). For a field →B=3.0×10−4e0.2x^j T, find the power required to move the conductor at a constant speed to x=2.0 m, y=0 m in 5×10–3 s. Assume parallel motion along the x-axis. (Assume e0.4=1.5)
- 14.85 W
- 29.7 W
- 1.57 W
- 4.5 W
Q. A loop of length l carrying current i is kept horizontally in a downward magnetic field B, the net force is?
- i(l x B)
- l(i.B)
- zero
- none of these
Q. Figure shows a square loop. 20 cm on each side in the x-y plane with its centre at the origin. The loop carries a current of 7 A.
Above it at y=0, z=12 cm is an infinitely long wire parallel to the x-axis carrying a current of 10 A. The net force on the loop is ———×10–5 N.
Above it at y=0, z=12 cm is an infinitely long wire parallel to the x-axis carrying a current of 10 A. The net force on the loop is ———×10–5 N.
Q. Two parallel wires aligned with y- axis carrying equal currents in opposite directions are placed at x=+a and x=−a in the x-y plane. The net magnetic field at the origin, O is B1 and at a point P (2a, 0, 0) is B2. Then the ratio B1:B2 is:
- 3:1
- 1:2
- 1:3
- 2:1