4 Marks Questions

Question 14 Marks

Predict the direction of induced current in the situations described by the following figures (a) to (f).

Answer

(a) As the magnet moves towards the solenoid, the magnetic flux associated with the solenoid increases. By Lenz's law, induced magnetic field (e.m.f.) opposes the cause producing it means it opposes the motion of the magnet. As a result, q becomes south pole and p acquires north polarity. Hence current will flow in the coil along the direction pq means current will flow along qrpq as shown in the figure. When seen from the side of the magnet, the direction will be clockwise as per clock rule. (b) As the north pole is moved away from the coil xy, magnetic flux associated with the coil also decrease. In this way according to Lenz's law the induced emf in the coil will oppose the motion of the magnet. Hence x becomes south pole so current will be in clockwise direction. The direction of current is along the angle yzx and for coil pq it is moving towards the end q and so this end will acquire south polarity so that it can oppose the motion of the magnet, hence current will flow in the coil along the direction prq. (c) Induced current will be in anticlockwise direction, along yzx. (d) Induced current will be in clockwise direction, along zyx. (e) In left coil, current in the battery will be from right towards left. Hence due to mutual inductance, induced current in the right coil will be in opposite direction which means from left to right or along xry. (f) No induced current because field lines are located in plane of the loop.

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Question 24 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

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 $

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