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
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.
A magnetised needle of magnetic moment 4.8 × 10−2 JT−1 is placed at 30° with the direction of uniform magnetic field of magnitude 3 × 10−2 T. Calculate the torque acting on the needle.
The rectangular wire-frame, shown in figure, has a width d, mass m, resistance R and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t = 0. (a) Find the acceleration of the frame when its speed has increased to v. (b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find this velocity v0. (c) Show that the velocity at time t is given by
v = v0(1 − e−Ft/mv0).
Find the magnetic field B at the centre of a rectangular loop of length l and width b, carrying a current i.
A rectangular conducting loop consists of two wires on two opposite sides of length l joined together by rods of length d. The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance R and the rods are of low resistance, which in turn are connected to a constant voltage source V0. The loop is placed in uniform a magnetic field B at 45° to its plane. Find τ, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.
- 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`.
