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.

Question

Philoid · Accepted Answer

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)