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Question
Two coaxial solenoids of different radii carry current I in the same direction. Let →F1 be the magnetic force on the inner solenoid due to the outer one and →F2 be the magnetic force on the outer solenoid due to the inner one. Then
Two coaxial solenoids of different radii carry current I in the same direction. Let →F1 be the magnetic force on the inner solenoid due to the outer one and →F2 be the magnetic force on the outer solenoid due to the inner one. Then
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Solution
The correct option is B →F1 = →F2 = 0
Assume axis of solenoid along x axis and z is in upward direction
Consider the force on inner solenoid, the magnetic field is constant along the axis of solenoid
The force on element 1 of solenoid coil is given as
dF=→il×→B
taking element on the coil
→il=il^j,→B=−B∘^idF=−ilB∘^j×^idF=ilB∘^k
the force will be in positive z direction
for diagonally opposite element 2 the magnitude of force will be same but the direction will be negative z direction.Force on element of any coil of inner solenoid will be radially inward which will cancel out and net result will be a zero force on coil, so value of
→F1=0
Now consider magnetic field of inner solenoid which is symmetric about it's axis.
The field is not exactly along the axis of outer solenoid.
The force on element of solenoid coil is given as
dF=→il×→B
taking element on the coil.
→il=il^j,→B=−B(^i+^k)dF=−ilB∘^j×(^k+^i)
The radial component of force will again be zero, but their will be a net pull or push on coil depending on position of coil of outer solenoid w.r.t. inner solenoid .
as the outer solenoid is placed symmetric the push of any coil will be equal and opposite to pull of coil placed exactly opposite distance from center. hence the net force on outer coil will be zero.
→F2=ilB^i−ilB^i=0
Assume axis of solenoid along x axis and z is in upward direction
Consider the force on inner solenoid, the magnetic field is constant along the axis of solenoid
The force on element 1 of solenoid coil is given as
dF=→il×→B
taking element on the coil
→il=il^j,→B=−B∘^idF=−ilB∘^j×^idF=ilB∘^k
the force will be in positive z direction
for diagonally opposite element 2 the magnitude of force will be same but the direction will be negative z direction.Force on element of any coil of inner solenoid will be radially inward which will cancel out and net result will be a zero force on coil, so value of
→F1=0
Now consider magnetic field of inner solenoid which is symmetric about it's axis.
The field is not exactly along the axis of outer solenoid.
The force on element of solenoid coil is given as
dF=→il×→B
taking element on the coil.
→il=il^j,→B=−B(^i+^k)dF=−ilB∘^j×(^k+^i)
The radial component of force will again be zero, but their will be a net pull or push on coil depending on position of coil of outer solenoid w.r.t. inner solenoid .
as the outer solenoid is placed symmetric the push of any coil will be equal and opposite to pull of coil placed exactly opposite distance from center. hence the net force on outer coil will be zero.
→F2=ilB^i−ilB^i=0
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