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A square loop of side a, resistance R, mass m is sliding as shown on a smooth horizontal table with speed v0. It enters a uniform magnetic field of magnitude B0 perpendicular to the table. It is seen that the loop comes to rest after entering a distance l inside the magnetic field. The value of v0 can be –
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A) B2a3mRB) B2a32mRC) B2a33mRD) B2a36mR

Answer
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Hint: We need to understand the factors that are affecting the motion of a resistor in a perpendicular magnetic field. We can relate the given quantities such as the magnetic field, resistance, and area of the loop to get the velocity of the square initially.

Complete step-by-step solution
We know that the magnetic field provides a force on a conducting material, which moves in the field with a perpendicular component of velocity to the direction of the magnetic field. We are given a square, which moves with its plane perpendicular to the direction of the magnetic field with an initial velocity.
The Lorentz force due to this motion experienced by the square loop is given by –
 F=BlI
Where B is the magnetic field strength,
 l is the side of the square loop,
‘I’ is the induced current in the square loop.
 
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We know that the current induced in a loop is given by the relation –
I=Bav0R
Substituting this in the equation for force gives –
I=Bav0RF=BaBav0RF=B2a2v0R
We can equate the force to the product of the mass and acceleration on the square loop as –
F=B2a2v0Rmaf=B2a2v0Rmdv0dt=B2a2v0Rbut,dv0dt=v0dv0dx
mv0dv0dx=B2a2v0Rmv0vdv0=B2a2R0xdxm(vv0)=B2a2xRv=v0B2a2xmRbut,v=0,x=av0=B2a3mR
This is the required initial velocity for the square to travel a distance of ‘l’ in the magnetic field.
The correct answer is option A.

Note: The moving coil or the square loop should have a perpendicular component along the direction of the magnetic field for any Lorentz force act upon the conducting coil or the resistor. We needn’t require the angle here because all the velocity was along the perpendicular.