Two coaxial solenoids of different radii carry current \(I\) in the same direction. Let \(\vec{F}_1\) be the magnetic force on the inner solenoid due to the outer one and \(\vec{F}_2\) be the magnetic force on the outer solenoid due to the inner one. Then:

1. \(\overrightarrow{{F}_1}=\overrightarrow{F_2}=0\)
2. \(\vec{F}_1\) is radially inwards and \(\vec{F}_2\) is radially outwards
3. \(\vec{F}_1\) is radially inwards and  \(\vec{F}_2=0\)
4. \(\vec{F}_1\) is radially outwards and  \(\vec{F}_2=0\)
Subtopic:  Force between Current Carrying Wires |
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Two long current carrying thin wires, both with current \(I\), are held by insulating threads of length \(L\) and are in equilibrium as shown in the figure, with threads making an angle '\(\theta\)' with the vertical. The mass per unit length of wires is \(\lambda\), then the value of \(I\) is:
(\(g=\) gravitational acceleration)
                        
1. \( \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}} \)
2. \( 2 \sin \theta \sqrt{\frac{\pi \lambda g L}{\mu_0 \cos \theta}} \)
3. \( 2 \sqrt{\frac{\pi g L}{\mu_0} \tan \theta} \)
4. \( 2 \sqrt{\frac{\pi \lambda g L}{\mu_0} \tan \theta}\)

Subtopic:  Force between Current Carrying Wires |
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Two long straight parallel wires, carrying (adjustable) current \({I_1}\) and \({I_2}\) are kept at a distance \({d}\) apart. If the force \(F\) between the two wires is taken as positive when the wires repel each other and negative when the wires attract each other, the graph showing the dependence of\(~F\), on the product \({I_1}{I_2}\) would be:
1. 3.
2. 4.
Subtopic:  Force between Current Carrying Wires |
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A rigid square loop of side \(a,\) carrying a current \(I_2,\) lies on a horizontal surface near a long, straight wire carrying a current \(I_1.\) Both the loop and the wire are in the same plane, as shown in the figure. The net force acting on the loop due to the current in the wire is:
         

1. zero
2. repulsive and equal to \( \dfrac{\mu_0 I_1 I_2}{4 \pi} \)
3. repulsive and equal to \( \dfrac{\mu_0 I_1 I_2}{2 \pi} \)
4. attractive and equal to \(\dfrac{\mu_0 I_1 I_2}{3 \pi}\)
Subtopic:  Force between Current Carrying Wires |
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Two wires \(A\) & \(B\) are carrying currents \(I_1\) & \(I_2\) as shown in the figure. The separation between them is \(d\). A third wire \(C\) carrying a current \(I\) is to be kept parallel to them at a distance \(x\) from \(A\) such that the net force acting on it is zero. The possible values of \(x\) are:

                   

1. \( x=\left(\frac{I_1}{I_1-I_2}\right) d \text { and } x=\frac{I_2}{\left(I_1-I_2\right)} d \)
2. \( x= \pm \frac{I_1 d}{\left(I_1-I_2\right)} \)
3. \( x=\left(\frac{I_1}{I_1-I_2}\right) d \text { and } x=\frac{I_2}{\left(I_1-I_2\right)} d \)
4. \( x=\left(\frac{I_2}{I_1-I_2}\right) d \text { and } x=\frac{I_2}{\left(I_1-I_2\right)} d\)
Subtopic:  Force between Current Carrying Wires |
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Two parallel, long wires are kept \(0.20~\text m\) apart in a vacuum, each carrying current of \(x ~\text A\) in the same direction. If the force of attraction per meter of each wire is \(2 \times 10^{-6} ~\text N,\) then the value of \(x \) is approximately:
1. \(1\)
2. \(2.4\)
3. \(1.4\)
4. \(2\)
Subtopic:  Force between Current Carrying Wires |
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A wire \(\mathrm X\) of length \(50~\text{cm}\) carrying a current \(2~\text{A}\) is placed parallel to a long wire \(\mathrm Y\) of length \(5~\text{m}\). The wire \(\mathrm Y\) carries a current \(3~\text{A}\). The distance between the two wires is \(5~\text{cm}\) and currents flow in the same direction. The force acting on the wire \(\mathrm{Y}\) is:
1. \(1.2\times 10^{-5}~\text{N}\) directed towards wire \(\mathrm{X}\).
2. \(1.2\times 10^{-4}~\text{N}\) directed away from wire \(\mathrm{X}\).
3. \(1.2\times10^{-4}~\text{N}\) directed towards wire \(\mathrm{X}\).
4. \(2.4\times 10^{-5}~\text{N}\) directed towards wire \(\mathrm{X}\).
Subtopic:  Force between Current Carrying Wires |
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Two parallel infinite wires carry equal currents as shown. If both the currents are doubled and separation is halved, the force on a \(10~\text{cm}\) section of one of the wires becomes:
                
1. \(4\) times 2. \(\dfrac{1}{4}\) times
3. \(8\) times 4. \(\dfrac{1}{8}\) times
Subtopic:  Force between Current Carrying Wires |
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