Advertisements
Advertisements
प्रश्न
A square loop of side 'a' carrying a current I2 is kept at distance x from an infinitely long straight wire carrying a current I1 as shown in the figure. Obtain the expression for the resultant force acting on the loop.
Advertisements
उत्तर
According to the right-hand screw rule, the magnetic field will be into the plane across the loop Force on length AD
F = Bil
`F_1 = (mu_0I_1I_2a)/(2pix)`
Force on length BC
F = Bil
`F_2 = (mu_0I_1I_2a)/(2pi(x + a))`
Force on AB and CD will be equal and opposite . Hence, they'll cancel out. Force on the loop
`F_"Net" = F_1 - F_2`
= `(mu_0I_1I_2a)/(2pi)[1/x - 1/((x + a))]`
`F_("Net") = (mu_0I_1I_2a)/(2pi)[(x + a - x)/((x + a)x)] = (mu_0I_1I_2a^2)/(2pi(x + a)x)`
`F_("Net") = (mu_0I_1I_2a^2)/(2pix(x + a))` (Towards left)
APPEARS IN
संबंधित प्रश्न
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s−1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.
(e = 1.5 × 10–19 C, me = 9.1 × 10–31 kg)
A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of 1 A. A straight long wire carrying 4 A current is kept near the loop as shown. If the loop and the wire are coplanar, find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.
A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1 A. A straight wire carrying 2 A current is kept near the loop as shown. If the loop and the wire are coplanar, find the (i) torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.
A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field `vecB`. Prove that the torque `vectau`acting on the loop is give by `vectau =vecm xx vecB,`where `vecm` is the magnetic moment of the loop.
Figure shows a square loop of edge a made of a uniform wire. A current i enters the loop at the point A and leaves it at the point C. Find the magnetic field at the point P which is on the perpendicular bisector of AB at a distance a/4 from it.
Find the magnetic field B at the centre of a rectangular loop of length l and width b, carrying a current i.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in fig.? What is the force on each case? Which case corresponds to stable equilibrium?
Consider the motion of a charged particle in a uniform magnetic field directed into the paper. If velocity v of the particle is in the plane of the paper the charged particle will ______.
- Assertion (A): The deflecting torque acting on a current-carrying loop is zero when its plane is perpendicular to the direction of the magnetic field.
- Reason (R): The deflecting torque acting on a loop of the magnetic moment `vecm` in a magnetic field `vecB` is given by the dot product of `vecm` and `vecB`.
