Question 14 Marks
A square loop of a metallic wire of side 10 cm whose resistance is 1 ohm is moving with a constant velocity of as shown in the figure in a uniform magnetic field of $200 Wb / m ^2$. The magnetic field is normal to the plane of paper inwards. This loop is connected to a network of resistors of magnitude 3 ohms. Resistance of wire OS and PQ are negligible. What will be the magnitude of velocity, $v_0$ to flow one ampere of direct current in the loop? Also, what will be the direction of current?
Answer
View full question & answer→It is clear from the figure that network of resistors is a balanced Wheatstone bridge. Hence, the resistor $3 \Omega$ in the side AC (diagonal) will not have any effect. QA and AS are in series among themselves. Let equivalent resistance of series combination is $R _1$.
$ \therefore R_1=3+3=6 \Omega $
Let equivalent resistance of QC and CS in series combination $= R _2$ (let) $=3+3=6 \Omega$.
Now, $R_1$ and $R_2$ are in parallel combination whose equivalent resistance $R ^{\prime}=\frac{ R _1 R _2}{ R _1+ R _2}=\frac{6 \times 6}{6+6}=\frac{36}{12}=3 \Omega$.
i.e., Resistance of loop is $1 \Omega$.
Hence, total resistance of the whole circuit $ =R^{\prime}+1=3+1=4 \Omega \text {. } $
Induced electromotive force due to motion of loop in magnetic field in loop $\varepsilon= Bv _0 I$
Current in the loop, $I =\frac{\varepsilon}{ R }=\frac{ Bv _0 l}{ R }$
$\therefore \quad$ Speed of loop $v_0=\frac{I \times R}{B \times l}$
Here, $I=1 mA=10^{-3} A, R =4 \Omega, B=2 Wb / m ^2, I =$ 0.1 m
On putting the values in (1)
$ v_0=\left(\frac{10^{-3} \times 4}{2 \times 0.1}\right)=2.0 \times 10^{-2} m / s \text { or } 2 cm / s $
$ \therefore R_1=3+3=6 \Omega $
Let equivalent resistance of QC and CS in series combination $= R _2$ (let) $=3+3=6 \Omega$.
Now, $R_1$ and $R_2$ are in parallel combination whose equivalent resistance $R ^{\prime}=\frac{ R _1 R _2}{ R _1+ R _2}=\frac{6 \times 6}{6+6}=\frac{36}{12}=3 \Omega$.
i.e., Resistance of loop is $1 \Omega$.
Hence, total resistance of the whole circuit $ =R^{\prime}+1=3+1=4 \Omega \text {. } $
Induced electromotive force due to motion of loop in magnetic field in loop $\varepsilon= Bv _0 I$
Current in the loop, $I =\frac{\varepsilon}{ R }=\frac{ Bv _0 l}{ R }$
$\therefore \quad$ Speed of loop $v_0=\frac{I \times R}{B \times l}$
Here, $I=1 mA=10^{-3} A, R =4 \Omega, B=2 Wb / m ^2, I =$ 0.1 m
On putting the values in (1)
$ v_0=\left(\frac{10^{-3} \times 4}{2 \times 0.1}\right)=2.0 \times 10^{-2} m / s \text { or } 2 cm / s $
