Question

A rectangular frame of wire abcd has dimensions 32 cm × 8.0 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 × 10⁻⁵ 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.

Answer 1

Given:
Total resistance of the frame, R = 2.0 Ω
Magnetic field, B = 0.020 T
Dimensions of the frame:
Length, l = 32 cm = 0.32 m
Breadth, b = 8 cm = 0.08 m

(a) Let the velocity of the frame be v.
The emf induced in the rectangular frame is given by
e = Blv
Current in the coil, $i=\frac{Blv}{R}$
The magnetic force on the rectangular frame is given by
F = ilB = 3.2 × 10⁻⁵ N
On putting the value of i, we get
$$\begin{gathered} \frac{B^2{l}^{2}v}{R}=3.2\times {10}^{-5} \\ \Rightarrow \frac{(0.020{)}^2\times (0.08{)}^2\times v}{2}=3.2\times {10}^{-5} \\ \Rightarrow v=\frac{3.2\times {10}^{-5}}{6.4\times {10}^{-3}\times 4\times {10}^{-4}} \\ =25\mathrm{m}/\mathrm{s} \end{gathered}$$

(b) Emf induced in the loop, e = vBl
⇒ e = 25 × 0.02 × 0.08
= 4 × 10⁻² V

(c) Resistance per unit length is given by
r$=\frac{2}{0.8}$
Ratio of the resistance of part, $\frac{ad}{cb}=\frac{2\times 0.72}{0.8}=1.8\mathrm{\Omega }$
$$\begin{gathered} V_{\mathrm{ab}}=iR=\frac{Blv}{2}\times 1.8 \\ =\frac{0.2\times 0.08\times 25\times 1.8}{2} \\ =0.036\mathrm{V}=3.6\times {10}^{-2}\mathrm{V} \end{gathered}$$

(d) Resistance of cd:

$$\begin{gathered} R_{\mathrm{cd}}=\frac{2\times 0.8}{0.8}=0.2\mathrm{\Omega } \\ V=i{R}_{\mathrm{cd}}=\frac{2\times 0.08\times 25\times 0.2}{2} \\ =4\times {10}^{-3}\mathrm{V} \end{gathered}$$