Q39 of 104 Page 306

A rectangular frame of wire abcd has dimensions 30 cm × 80 cm and a total resistance of 2.0 Ω. It is pulled out of a magnetic field B = 0.020 T by applying a force of 3.2 × 10–6 N (figure). It is found that the frame moves with constant speed. Find

(a) this constant speed,


(b) the emf induced in the loop,


(c) the potential difference between the points a and b and (d) the potential difference between the points c and d.



Given:


Length ab = cd = 30 cm = 0.3 m


Length bc = ad = 80 cm = 0.8 m


Total resistance R = 2 Ω


Magnetic field B = 0.02 T


Force F = 3.2 × 10–6 N


Formula used:


(a) Magnetic force on a current carrying wire … (i),where I = current, l = length of wire, B = magnetic field.


Hence, current … (ii)


Now, emf … (iii), where B = magnetic field, l = length of wire, v = constant velocity with which it is moving


Also, by Ohm’s law, … (iv), where I = current,


R = resistance.


Hence, equating (iii) and (iv) and substituting I from (ii), we get


, where B = magnetic field, v = velocity, F = force, R = resistance, l = length of wire.


Here, since the force is applied on the side cd, we consider


l =30 cm = 0.3 m (the shorter length).


Hence,


Substituting the given values, we get


ms-1 = 0.18 ms-1


Constant speed with which the frame moves = 0.18 ms-1(Ans)


(b) Emf induced in the loop , where B = magnetic field, l = length of the wire which is moving, v = velocity


Hence, = 0.001 V


Emf induced in loop = 0.001 V (Ans)


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37

Consider the situation of the previous problem.

(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.


 38

Consider the situation shown in figure. The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separated by a distance ℓ. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangular region and a resistance R connects the rails outside the field region. At t = 0, the wire PQ is pushed towards right with a speed v0. Find

(a) the current in the loop at an instant when the speed of the wire PQ is v,


(b) the acceleration of the wire at this instant,


(c) the velocity v as a function of x and


(d) the maximum distance the wire will move.



 40

Figure shows a metallic wire of resistance 0.20 Ω sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 μA passes through the wire when it is slid at a rate of 20 cm s–1. If the horizontal component of the earth’s magnetic field is 3.0 × 10–5 T, calculate the dip at the place.


 41

A wire ab of length ℓ, mass m and resistance R slides on a smooth, thick pair of metallic rails joined at the bottom as shown in figure. The plane of the rails makes an angle θ with the horizontal. A vertical magnetic field B exists in the region. If the wire slides on the rails at a constant speed v, show that