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CBSE Class 12 Physics 2016 Delhi Set 1 Paper

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Question : 16 of 26
Marks: +1, -0
Two long straight parallel conductors carry steady current I1I_1 and I2I_2 separated by a distance dd. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.
Solution:  
Diagram showing attractive force on other wire.
Obtaining an expression for force.
Definition of one ampere.
As shown in Figure, the direction of force on conductor bb is attractive
[Alternatively:
B\overset{\rightarrow}{B} at a point on wire 2 , is along k^-\hat{k}
F\therefore \overset{\rightarrow}{F}, on wire 2 , due to the B\overset{\rightarrow}{B}, is along i^-\hat{i}, i.e. towards wire1. Hence the force is attractive.
Magnetic field, due to current in conductor aa ,
B1=μ0I12πdB_1 = \frac{\mu_0 I_1}{2\pi d}
The magnitude of force on a length LL of conductor bb ,
F2=I2LB1F_2 = I_2 L B_1
F2=μ0I1I2L2πdF_2 = \frac{\mu_0 I_1 I_2 L}{2\pi d}
One ampere is that steady current which, when maintained in each of the two very long, straight, parallel conductors, placed one meter apart in vacuum, would produce on each of these conductors a force equal to 2×1072 \times 10^{-7} newton per meter of theirlength.
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