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# 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 - Physics

ConceptTorque on Current Loop, Magnetic Dipole Torque on a Rectangular Current Loop in a Uniform Magnetic Field

#### Question

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.

#### Solution

(1) vectau = vecM xx vecB = MBsintheta

Here M and B have the same direction

θ = 0°

|vectau| = MBsin theta =0

(2) We know vecF_B = i vecI xx vecB

On line AB, and CD magnetic forces are equal and opposite. So they cancel out each other. Magnetic force on line AD.

vecF = ivecl xx vecB   (Attractive)

=ilB                      [l= 4cm = 0.04m]

because B=(mu_0I) /(2pir)

|vecF| = (muiIl)/(2pir) (Attractive)

Magnetic force on line CB.

vecF = ivecl xx vecB'   (Repulsive)

=> F'=|vecF| = ilB'

because B' = (mu_0I)/(2pir')

F' = (mu_0iIl)/(2pir')  (Repulsive)

So, net force

F_n = F -F'

=(mu_0iIl)/(2pi) [1/r - 1/(r')]

Given

I = 2A

I = 1 A

= 2 cm = 0.02 m

r′ = (2 + 2.5) cm = 4.5 cm

= 0.045 m

l = 4 cm = 0.04 m

Plug-in these values in above equation

F_n = ((4pi xx 10_7)(2)(1)(0.04))/(2pi)   [1/0.02 - 1/0.045]

=  4 xx 10^-7 xx  [1/2 - 1/4.5]

= 16 xx 10^-7[0.5 -0.22]

= 16 xx 10^-7 [0.278]

= 4.45 xx 10^-7 N

The direction of the force on the loop will be towards left.

Is there an error in this question or solution?

#### APPEARS IN

Solution 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 Concept: Torque on Current Loop, Magnetic Dipole - Torque on a Rectangular Current Loop in a Uniform Magnetic Field.
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