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Step 1: Understand the Physical Situation
A ring of mass 4\,\text{kg} with uniform linear charge density \lambda = 4\,\text{C/m} is placed on a rough horizontal surface, where the coefficient of friction is \mu = \frac{\pi}{4}. A time-varying magnetic field given by
B(t) = B_{0} \, t^{2}
is applied perpendicular to the plane of the ring in a circular region of radius a = 5\,\text{cm}. We need to find the time when the ring just begins to rotate, overcoming the frictional torque.
Step 2: Convert and Note All Given Quantities
1. Radius of the circular region:
a = 5\,\text{cm} = 0.05\,\text{m}.
2. Mass of the ring:
m = 4\,\text{kg}.
3. Linear charge density:
\lambda = 4\,\text{C/m}.
4. Friction coefficient:
\mu = \frac{\pi}{4}.
5. Acceleration due to gravity:
g = 10\,\text{m/s}^{2}.
6. Magnetic field coefficient:
B_{0} = 125\,\text{T/s}^{2}.
Step 3: Write the Magnetic Flux and Its Rate of Change
1. The magnetic flux through an area \pi a^{2} (where a is the radius of the circular region) is
\Phi = B(t) \times \text{Area}
= \bigl(B_{0}\,t^{2}\bigr)\,\pi a^{2}.
2. Its time derivative (rate of change of flux) is
\frac{d\Phi}{dt}
= \frac{d}{dt}\bigl(B_{0}\,\pi\,a^{2}\,t^{2}\bigr)
= 2\,B_{0}\,\pi\,a^{2}\,t.
Step 4: Induced Electric Field Around the Ring
From Faraday’s law in integral form for a circular path of radius a ,
E \cdot (2\pi a) = \left|\frac{d\Phi}{dt}\right| = 2\,B_{0}\,\pi\,a^{2}\,t,
so the magnitude of the induced electric field is
E
= \frac{2\,B_{0}\,\pi\,a^{2}\,t}{2\pi a}
= B_{0}\,a\,t.
Step 5: Total Charge on the Ring and the Tangential Force
1. Circumference of the ring is 2\pi a .
2. Total charge on the ring:
Q = \lambda \times (2\pi a).
3. Tangential force on the ring due to the induced electric field:
F = Q\,E
= \bigl[\lambda \,(2\pi a)\bigr]\bigl(B_{0}\,a\,t\bigr).
Step 6: Torque Induced by the Electric Field
The ring’s radius is a , so the torque due to this tangential force is
\tau_{1} = F \times a
= \bigl[\lambda \,(2\pi a)\bigr]\bigl(B_{0}\,a\,t\bigr)\,a
= 2\pi\,\lambda\,B_{0}\,a^{3}\,t.
Step 7: Frictional Torque Opposing the Motion
1. Maximum frictional force on the ring:
F_{\text{friction}} = \mu \, m \, g.
2. This force acts at the outer edge (radius a ), hence the frictional torque is
\tau_{\text{fric}}
= F_{\text{friction}} \times a
= (\mu\,m\,g)\,a.
Step 8: Condition for the Ring to Start Rotating
The ring just begins to rotate when the induced torque equals the frictional torque:
\tau_{1} = \tau_{\text{fric}},
so
2\pi\,\lambda\,B_{0}\,a^{3}\,t
= \mu\,m\,g\,a.
Solving for t :
t
= \frac{\mu\,m\,g\,a}{2\pi\,\lambda\,B_{0}\,a^{3}}
= \frac{\mu\,m\,g}{2\pi\,\lambda\,B_{0}\,a^{2}}.
Step 9: Numerical Substitution
Substitute the given values:
\mu = \frac{\pi}{4}
m = 4\,\text{kg}
g = 10\,\text{m/s}^{2}
a = 0.05\,\text{m}
\lambda = 4\,\text{C/m}
B_{0} = 125\,\text{T/s}^{2}
Thus,
t
= \frac{\left(\frac{\pi}{4}\right) \times 4 \times 10}{2\pi \times 4 \times 125 \times (0.05)^{2}}
= 4\,\text{s}.
Final Answer: The ring just starts rotating after 4\,\text{seconds}.
