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Home » We have a square loop having side as 12 cm and its sides are parallel to x and y-axis is moved with a velocity of 8 cm /s in the positive x-direction in a region which have a magnetic field in the direction of positive z-axis. The field is not uniform whether in case of its space or in the case of time. It has a gradient of 10−3 T cm−1 along the negative x-direction(i.e its value increases by 10−3 T cm−1 as we move from positive to negative direction ), and it is reducing in the case of time with the rate of 10−3 T s−1 . Calculate the magnitude and direction of induced current in the loop (Given: Resistance = 4.50mΩ).
We have a square loop having side as 12 cm and its sides are parallel to x and y-axis is moved with a velocity of 8 cm /s in the positive x-direction in a region which have a magnetic field in the direction of positive z-axis. The field is not uniform whether in case of its space or in the case of time. It has a gradient of 10−3 T cm−1 along the negative x-direction(i.e its value increases by 10−3 T cm−1 as we move from positive to negative direction ), and it is reducing in the case of time with the rate of 10−3 T s−1 . Calculate the magnitude and direction of induced current in the loop (Given: Resistance = 4.50mΩ).
CBSE | Class 12 | Electromagnetic Induction | NCERT | Physics
We have a square loop having side as 12 cm and its sides are parallel to x and y-axis is moved with a velocity of 8 cm /s in the positive x-direction in a region which have a magnetic field in the direction of positive z-axis. The field is not uniform whether in case of its space or in the case of time. It has a gradient of 10−3 T cm−1 along the negative x-direction(i.e its value increases by 10−3 T cm−1 as we move from positive to negative direction ), and it is reducing in the case of time with the rate of 10−3 T s−1 . Calculate the magnitude and direction of induced current in the loop (Given: Resistance = 4.50mΩ).

Answer –

We have,

Side of the Square loop is given by s = 12cm = 0.12m

Area of the loop becomes –

 A = s × s = 0.12 × 0.12

A = 0.0144 m2

Velocity of the loop is given by v = 8 cm-1 = 0.08 ms-1

Expression of the gradient of the magnetic field along negative x-direction is –

    \[\frac{dB}{dx}={{10}^{-3}}Tc{{m}^{-1}}={{10}^{-1}}T{{m}^{-1}}\]

And, the rate of decrease of the magnetic field is given by –

    \[\frac{dB}{dt}={{10}^{-3}}T{{s}^{-1}}\]

Resistance R is given = 4.50mΩ

In a non-uniform magnetic field, the rate of change of the magnetic flux due to loop motion is given as –

    \[\frac{dB}{dt}=A\times \frac{dB}{dx}\times v=144\times {{10}^{-4}}\times {{10}^{-1}}\times 0.08\]

    \[\frac{dB}{dt}=11.52\times {{10}^{-5}}T{{m}^{2}}{{s}^{-1}}\]

Rate of change of the flux due to explicit time variation in field B is given by –

    \[\frac{d\phi }{dt}=A\frac{dB}{dt}=144\times {{10}^{-4}}\times {{10}^{-3}}\]

    \[\frac{d\phi }{dt}=1.44\times {{10}^{-5}}T{{m}^{2}}{{s}^{-1}}\]

Because the induced emf is the rate of change of the flux, the total induced emf in the loop may be calculated as –

    \[e=1.44\times {{10}^{-5}}+11.52\times {{10}^{-5}}\]

    \[e=12.96\times {{10}^{-5}}V\]

Now, the induced current is given by –

    \[i=\frac{e}{R}=\frac{12.96\times {{10}^{-5}}}{4.5\times {{10}^{-3}}}\]

    \[i=2.88\times {{10}^{-2}}A\]

As a result, the induced current is directed in such a way that the flux through the loop increases in the positive z-direction.

i

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