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Figure Shows a Long U-shaped Wire of Width L Placed in a Perpendicular Magnetic Field B. - Physics

Sum

Figure shows a long U-shaped wire of width l placed in a perpendicular magnetic field B. A wire of length l is slid on the U-shaped wire with a constant velocity v towards right. The resistance of all the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. (a) Calculate the force needed to keep the sliding wire moving with a constant velocity v. (b) If the force needed just after t = 0 is F0, find the time at which the force needed will be F0/2.0

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Solution

Emf induced in the circuit, e = Bvl

Current in the circuit,

\[i = \frac{e}{R} = \frac{Bvl}{2r(l + vt)}\]

(a) Force F needed to keep the sliding wire moving with a constant velocity v will be equal in magnitude to the magnetic force on it. The direction of force F will be along the direction of motion of the sliding wire.

Thus, the magnitude of force F is given by

\[F = ilB = \frac{Bvl}{2r(l + vt)} \times lB\]

\[ = \frac{B^2 l^2 v}{2r(l + vt)}\]

(b) The magnitude of force F at t = 0 is given by

\[F_0 = ilB = lB\left( \frac{lBv}{2rl} \right) \]

\[ = \frac{l B^2 v}{2r} .............(1)\]

Let at time t = T,  the value of the force be F0/2.

Now,

\[\frac{F_0}{2} = \frac{l^2 B^2 v}{2r(l + vT)}\]

On substituting the value of F0 from (1), we get

\[\frac{l B^2 v}{4r} = \frac{l^2 B^2 v}{2r(l + vT)}\]

\[ \Rightarrow 2l = l + vT\]

\[ \Rightarrow T = \frac{l}{v}\]

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 16 Electromagnetic Induction
Q 37 | Page 308
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