A rectangular loop sides 10cm and 3cm moving out of a region of uniform magnetic field of 0.5T directed normal to the loop if we want to move loop with a constant velocity 1cm/sec. Then required mechanical force is (R=1mohm)?
physics - EMI
Answers
The required mechanical force is 2.25 * 10⁻³ N.
Explanation:
=> It is given that,
Uniform magnetic field of rectangular loop, B = 0.5 T
length of a side of a rectangular loop, l = 3 cm = 0.03 m
Constant Velocity of loop, v = 1 cm/ sec
Resistance of loop = 1 m ohm = 1 * 10⁻³
=> Suppose, a given rectangular loop is ABCD.
Area of loop = lx = ΔS
=> the magnetic flux linked by loop:
flux = Blx = ∅
∴ d∅/dt = d(Blx)/dt =Blv
=> developed emf, e:
e = - d∅/dt = Blv [∵ x is decreasing]
=> Thus, the current in the loop, I:
I = e / R
= Blv / R
=> Required mechanical force, F:
F = B*I*L
= B²l²L/R
= (0.5)² * (0.03)² * 1 / 1 * 10⁻³
= 2.25 * 10⁻³ N
Thus, the required mechanical force is 2.25 * 10⁻³ N.
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