A wire carrying current I is tied between points P and Q and is in the shape of a circular arc of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle $2{\theta _0}$ at the center of the circle (of which it forms an arc) then the tension in the wire is:
A) $IBR$
B) $\dfrac{{IBR}}{{\sin {\theta _0}}}$
C) $\dfrac{{IBR}}{{2\sin {\theta _0}}}$
D) $\dfrac{{IBR{\theta _0}}}{{\sin {\theta _0}}}$
Answer
603.3k+ views
Hint:
Recall that tension is described as the pulling force acting by the means of a rope, string or a cable. It can also be written as the pair of action-reaction forces acting at the end of the given rope or string or cable. It is the opposite of compression.
Step-by-Step Explanation:
Step I:
Suppose the length of the wire is divided into very small elements, each of which is subtending an angle $d\theta $ at the origin of the wire. The magnetic force will act in an outward direction. Since the tension is also considered an action reaction force, it should balance the magnetic force. Then the wire is said to be in equilibrium.
Step II:
The magnetic force on each element of the wire is given by $F = Idl \times B$
Where I is the current
dl is the length
B is the magnetic field.
Step III:
The tension force will act in downward direction and balance the magnetic force. The resultant force will be equal to the component of tension and is given by
$F = {T_{resul\tan t}}$ ---(i)
${T_{(resultant)}} = 2T\sin \dfrac{{d\theta }}{2}$
And force is given by $F = IdlB$
Since $d\theta $ is a very small angle, therefore $\sin \dfrac{{d\theta }}{2} = \dfrac{{d\theta }}{2}$
And $d\theta = \dfrac{{dl}}{R}$ where R is the radius of the arc.
Step IV:
Substituting value of F and T in equation (i)
$IdlB = 2T\dfrac{{dl}}{{2R}}$
$IdlB = T\dfrac{{dl}}{R}$
$T = \dfrac{{IdlBR}}{{dl}}$
$T = IBR$
Step V:
The tension in the wire is $IBR$
Option A is the right answer.
Note:It is to be noted that the direction of magnetic field due to straight and circular loops are different. At the centre of the circular loop, the magnetic field lines are straight. Each segment of circular loop carrying current produces magnetic field lines in the same direction within the loop. The direction of magnetic field at the centre of circular coil is perpendicular to the place of the coil.
Recall that tension is described as the pulling force acting by the means of a rope, string or a cable. It can also be written as the pair of action-reaction forces acting at the end of the given rope or string or cable. It is the opposite of compression.
Step-by-Step Explanation:
Step I:
Suppose the length of the wire is divided into very small elements, each of which is subtending an angle $d\theta $ at the origin of the wire. The magnetic force will act in an outward direction. Since the tension is also considered an action reaction force, it should balance the magnetic force. Then the wire is said to be in equilibrium.
Step II:
The magnetic force on each element of the wire is given by $F = Idl \times B$
Where I is the current
dl is the length
B is the magnetic field.
Step III:
The tension force will act in downward direction and balance the magnetic force. The resultant force will be equal to the component of tension and is given by
$F = {T_{resul\tan t}}$ ---(i)
${T_{(resultant)}} = 2T\sin \dfrac{{d\theta }}{2}$
And force is given by $F = IdlB$
Since $d\theta $ is a very small angle, therefore $\sin \dfrac{{d\theta }}{2} = \dfrac{{d\theta }}{2}$
And $d\theta = \dfrac{{dl}}{R}$ where R is the radius of the arc.
Step IV:
Substituting value of F and T in equation (i)
$IdlB = 2T\dfrac{{dl}}{{2R}}$
$IdlB = T\dfrac{{dl}}{R}$
$T = \dfrac{{IdlBR}}{{dl}}$
$T = IBR$
Step V:
The tension in the wire is $IBR$
Option A is the right answer.
Note:It is to be noted that the direction of magnetic field due to straight and circular loops are different. At the centre of the circular loop, the magnetic field lines are straight. Each segment of circular loop carrying current produces magnetic field lines in the same direction within the loop. The direction of magnetic field at the centre of circular coil is perpendicular to the place of the coil.
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