Force on Current Carrying Conductor Class 12 Notes | EduRev

Physics Class 12

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Class 12 : Force on Current Carrying Conductor Class 12 Notes | EduRev

The document Force on Current Carrying Conductor Class 12 Notes | EduRev is a part of the Class 12 Course Physics Class 12.
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3. FIELD DUE TO A STRAIGHT CURRENT CARRYING WIRE

3.1 WHEN THE WIRE IS OF FINITE LENGTH

Consider a straight wire segment carrying a current i and there is a point P at which magnetic field to be calculated as shown in the figure. This wire segment makes angle θ1 and θ2 at that point with normal OP. Consider an element of length dy at a distance y from O and distance of this element from point P is r and line joining P to Q makes an angle q with the direction of current as shown in figure. Using Biot-Savart Law magnetic field at point P due to small current element is given by

Force on Current Carrying Conductor Class 12 Notes | EduRev

As every element of the wire contributes to Force on Current Carrying Conductor Class 12 Notes | EduRev in the same direction, we have 
Force on Current Carrying Conductor Class 12 Notes | EduRev

Force on Current Carrying Conductor Class 12 Notes | EduRev ....(i)

From the triangle OPQ as shown in diagram, we have

y = d tan Force on Current Carrying Conductor Class 12 Notes | EduRev

or dy = d sec2Force on Current Carrying Conductor Class 12 Notes | EduRevdForce on Current Carrying Conductor Class 12 Notes | EduRev

and is same triangle,

r = d sec Force on Current Carrying Conductor Class 12 Notes | EduRevand q = (90º - Force on Current Carrying Conductor Class 12 Notes | EduRev), where Force on Current Carrying Conductor Class 12 Notes | EduRevis angle between line OP and PQ

Now equation (i) can be written in this form

Force on Current Carrying Conductor Class 12 Notes | EduRev

or Force on Current Carrying Conductor Class 12 Notes | EduRev ...(3)

Note : θ1 & θ2 must be taken with sign
Force on Current Carrying Conductor Class 12 Notes | EduRev

For the case shown in figure

B at A = Force on Current Carrying Conductor Class 12 Notes | EduRev

Direction of Force on Current Carrying Conductor Class 12 Notes | EduRevThe direction of magnetic field is determined by the cross product of the vector Force on Current Carrying Conductor Class 12 Notes | EduRev with Force on Current Carrying Conductor Class 12 Notes | EduRev. Therefore, at point P, the direction of the magnetic field due to the whole conductor will be perpendicular to the plane of paper and going into the plane.

Right-hand Thumb Rule : The direction of B at a point P due to a long, straight wire can be found by the right-hand thumb rule. The direction of magnetic field is perpendicular to the plane containing wire and perpendicular from the point. The orientation of magnetic field is given by the direction of curl fingers if we stretch thumb along the wire in the direction of current. Refer figure.

Force on Current Carrying Conductor Class 12 Notes | EduRev Force on Current Carrying Conductor Class 12 Notes | EduRev

Conventionally, the direction of the field perpendicular to the plane of the paper is represented by Force on Current Carrying Conductor Class 12 Notes | EduRev if into the page and by Force on Current Carrying Conductor Class 12 Notes | EduRevif out of the page.

Now consider some special cases involving the application of equation (3)

Case 1 : When the point P is on the perpendicular bisector

In this case angle θ1 = θ2, using result of equation (3), the magnetic field is

Force on Current Carrying Conductor Class 12 Notes | EduRev

Force on Current Carrying Conductor Class 12 Notes | EduRev

where Force on Current Carrying Conductor Class 12 Notes | EduRev

Case - 2

(i) If the wire is infinitely long then the magnetic field at `P' (as shown in the figure) is given by (using q1 = q2 = 90° and the formula `B' due to straight wire)

B = Force on Current Carrying Conductor Class 12 Notes | EduRev ⇒ B µ Force on Current Carrying Conductor Class 12 Notes | EduRev   Force on Current Carrying Conductor Class 12 Notes | EduRev

The direction of Force on Current Carrying Conductor Class 12 Notes | EduRev at various is as shown in the figure. The magnetic lines of force will be concentric circles around the wire (as shown earlier)

(ii) If the wire is infinitely long but `P' is as shown in the figure. The direction of Force on Current Carrying Conductor Class 12 Notes | EduRevat various points is as shown in the figure. At `P'

B = Force on Current Carrying Conductor Class 12 Notes | EduRev

Force on Current Carrying Conductor Class 12 Notes | EduRev

Case III : When the point lies along the length of wire (but not on it)

Force on Current Carrying Conductor Class 12 Notes | EduRev

If the point P is along the length of the wire (but not one it), then as Force on Current Carrying Conductor Class 12 Notes | EduRev and Force on Current Carrying Conductor Class 12 Notes | EduRev will either be parallel or antiparallel, i.e., q = 0 or p, so Force on Current Carrying Conductor Class 12 Notes | EduRevand hence using equation (1)

Force on Current Carrying Conductor Class 12 Notes | EduRev

Ex-1 Calculate the magnetic field induction at a point distance, Force on Current Carrying Conductor Class 12 Notes | EduRev metre from a straight wire of length `a' metre carrying a current of i amp. The point is on the perpendicular bisector of the wire.

Sol. B = Force on Current Carrying Conductor Class 12 Notes | EduRev [sinq1 + sinq2]

= 10-7Force on Current Carrying Conductor Class 12 Notes | EduRev                      Force on Current Carrying Conductor Class 12 Notes | EduRev

Force on Current Carrying Conductor Class 12 Notes | EduRev

Perpendicular to the plane of figure (inward).

Ex.2 Find resultant magnetic field at `C' in the figure shown.

Force on Current Carrying Conductor Class 12 Notes | EduRev

Sol. It is clear that `B' at `C' due all the wires is directed Ä. Also B at `C due PQ and SR is same. Also due to QR and PS is same

Therefore, Bres = 2(BPQ + BSP)

BPQ = (sin 60° + sin 60°)

BSP Force on Current Carrying Conductor Class 12 Notes | EduRev (sin 30° + sin 30°)

⇒ BresForce on Current Carrying Conductor Class 12 Notes | EduRev = Force on Current Carrying Conductor Class 12 Notes | EduRev

Ex.3 Figure shows a square loop made from a uniform wire. Find the magnetic field at the centre of the square if a battery is connected between the points A and C.

Force on Current Carrying Conductor Class 12 Notes | EduRev

Sol. The current will be equally divided at A. The fields at the centre due to the currents in the wires AB and DC will be equal in magnitude and opposite in direction. The resultant of these two fields will be zero. Similarly, the resultant of the fields due to the wires AD and BC will be zero. Hence, the net field at the centre will be zero.

Ex.4 In the figure shown there are two parallel long wires (placed in the plane of paper) are carrying currents 2 I and I consider points A, C, D on the line perpendicular to both the wires and also in the plane of the paper. The distances are mentioned.

Force on Current Carrying Conductor Class 12 Notes | EduRev

Find (i) Force on Current Carrying Conductor Class 12 Notes | EduRev at A, C, D

(ii) position of point on line A C D where Force on Current Carrying Conductor Class 12 Notes | EduRev is zero.

Sol. (i) Let us call Force on Current Carrying Conductor Class 12 Notes | EduRev due to (1) and (2) as Force on Current Carrying Conductor Class 12 Notes | EduRev and Force on Current Carrying Conductor Class 12 Notes | EduRev respectively. Then

at A : Force on Current Carrying Conductor Class 12 Notes | EduRev is Force on Current Carrying Conductor Class 12 Notes | EduRev and Force on Current Carrying Conductor Class 12 Notes | EduRev is Force on Current Carrying Conductor Class 12 Notes | EduRev

B1Force on Current Carrying Conductor Class 12 Notes | EduRev and B2Force on Current Carrying Conductor Class 12 Notes | EduRev

Therefore, Bres = B1 - B2Force on Current Carrying Conductor Class 12 Notes | EduRev Force on Current Carrying Conductor Class 12 Notes | EduRev Ans.

at C: Force on Current Carrying Conductor Class 12 Notes | EduRev is Force on Current Carrying Conductor Class 12 Notes | EduRev and Force on Current Carrying Conductor Class 12 Notes | EduRev also Force on Current Carrying Conductor Class 12 Notes | EduRev

Therefore, Bres = B1 + B2

Force on Current Carrying Conductor Class 12 Notes | EduRev + Force on Current Carrying Conductor Class 12 Notes | EduRev = Force on Current Carrying Conductor Class 12 Notes | EduRev = Force on Current Carrying Conductor Class 12 Notes | EduRev Force on Current Carrying Conductor Class 12 Notes | EduRev Ans.

Force on Current Carrying Conductor Class 12 Notes | EduRev
Therefore, Bres = 0 Ans.

(ii) It is clear from the above solution that B = 0 at point `D'.

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