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प्रश्न
Write the expression for the magnetic moment `vecm`due to a planar square loop of side ‘l’ carrying a steady current I in a vector form.
In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current I1 at a distance l as shown. Give reason to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop.
उत्तर
The expression for the magnetic moment `vecm`due to a planar square loop of side ‘l’ carrying a steady current I in a vector form is given as
`vecm = IvecA`
Therefore,
`vecm = I(l)^2 hatn`
Where, `vecn` is the unit vector along the normal to the surface of the loop.
The attractive force per unit length on the loop is
`F_a = (mu_0)/(2pi) (I_1I)/I`
`F_a = (mu_0)/(2pi)I_1I`
The repulsive force per unit length on the loop is
`F_r/l = (mu _0)/(2pi) (I_1I)/(2l)`
`F_r = (mu _0)/(2pi) (I_1I)/(2l)`
`F_(\text { net}) = F_a - F_t`
= `(mu_0 )/(2pi)I_1I(1-1/2)`
`|F_(text {net}) = (mu_0)/(4pi)I_1I|`
Since the attractive force is greater than the repulsive force, a net force acts on the loop.
The torque on the loop is given as
`tau = vecm xx vecB`
`=mBsintheta`
`=IAB sintheta`
θ = 0° (`because`Area vector is parallel to the magnetic field)
τ = IAB sin0°
τ = 0
∴ The torque acting on the loop is zero.
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