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प्रश्न
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 × 10−5N (see the following 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 aand b and (d) the potential difference between the points c and d.
उत्तर
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
\[\frac{B^2 l^2 v}{R} = 3 . 2 \times {10}^{- 5} \]
\[ \Rightarrow \frac{(0 . 020 )^2 \times (0 . 08 )^2 \times v}{2} = 3 . 2 \times {10}^{- 5} \]
\[ \Rightarrow v = \frac{3 . 2 \times {10}^{- 5}}{6 . 4 \times {10}^{- 3} \times 4 \times {10}^{- 4}}\]
\[ = 25 m/s\]
(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=2/0.8`
Ratio of the resistance of part,
\[\frac{ad}{cb} = \frac{2 \times 0 . 72}{0 . 8} = 1 . 8 \Omega\]
\[V_{ab} = iR = \frac{Blv}{2} \times 1 . 8\]
\[ = \frac{0 . 2 \times 0 . 08 \times 25 \times 1 . 8}{2}\]
\[ = 0 . 036 V = 3 . 6 \times {10}^{- 2} V\]
(d) Resistance of cd:-
\[R_{cd} = \frac{2 \times 0 . 8}{0 . 8} = 0 . 2 \Omega\]
\[V = i R_{cd} = \frac{2 \times 0 . 08 \times 25 \times 0 . 2}{2}\]
\[= 4 \times {10}^{- 3} V\]
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