Find magnetic field at point P shown in Fig. point P is on the bisector of angle between the wires.

Find magnetic field at point P shown in Fig. point P is on the bisecto

1.	Find magnetic field at point P shown in Fig. point P is on the bisector of angle between the wires.
Answer» Assume `PO=x, so PQ=xsin(alpha//2)` Magnetic field B at P due to either segment of wire is `B=(mu_0I)/(4pix sin(alpha//2))[1+cosalpha/2]` Net magnetic field at P is `B_("net")=(mu_0I)/(2pix sin(alpha//2))[1+cos(alpha//2)]`

Discussion

No Comment Found

Related InterviewSolutions

Consider six wires coming into or out of the page, all with the same current. Rank the line integral of the magnetic field (from most positive to most negative) taken counterclockwise around each loop shown A. `BgtCgtDgtA`B. `BgtC=DgtA`C. `BltAgtC=D`D. `CltB=DgtA`
Two positive charges `q_1 and q_2` are moving with velocities `v_1 and v_2` when they are at points A and B, respectively, as shown in Fig. The magnetic force experienced by charge `q_1 ` due to the other charge `q_2` is A. `(mu_0q_1q_2v_1v_2)/(8sqrt2pia^2)`B. `(mu_0q_1q_2v_1v_2)/(4sqrt2pia^2)`C. `(mu_0q_1q_2v_1v_2)/(2sqrt2pia^2)`D. `(mu_0q_1q_2v_1v_2)/(sqrt2pia^2)`
A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be `(1)/(8)` th to its value at the centre of the coil, isA. `(R)/(sqrt(3))`B. `Rsqrt(3)`C. `2sqrt(3)R`D. `(2)/(sqrt(3))R`
An otherwise infinite, straight wire has two concentric loops of radii a and b carrying equal currents in opposite directions as shown in Fig. The magnetic field at the common centre is zero for A. `a/b=(pi-1)/pi`B. `a/b=pi/(pi+1)`C. `a/b=(pi-1)/(pi+1)`D. `a/b=(pi+1)/(pi-1)`
`L` is a circular ring made of a uniform wire, currents enters and leaves the ring through straight conductors which, if produces, would have passed through the centre `C` of ring. The magnetic field at `C` (i) due to the straight conductors is zero (ii) due to the loop is zero (iii) due to the loop is proportional to `theta` (iv) due to loop is proportional to `(pi-theta)`A. due to the straight conductors is zeroB. due to the loop is zeroC. due to path II is proportional to `theta`.D. due to the path I is proportional `(pi-theta)`
A long insulated copper wire is closely wound as a spiral of `N` turns. The spiral has inner radius a and outer radius `b`. The spiral lies in the `xy`-plane and a steady current I flows through the wire. The`z`-component of the magetic field at the centre of the spiral is A. `(mu_0NI)/(2(b-a)) In (b/a)`B. `(mu_0NI)/(2(b-a)) In ((b+a)/(b-a))`C. `(mu_0NI)/(2b) In (b/a)`D. `(mu_0NI)/(2b) In ((b+a)/(b-a))`
Four parallel conductors, carrying equal currents, pass vertically through the four corners of a square WXYZ. In two conductors, the current is flowing into the page, and in the other two out into the page, and in the other two out of the page. In what directions must the currents flow to produce a resultant magnetic field in the direction shown at O, the centre of the square? A. W and Y X and ZB. X and Z W and YC. W and Z X and YD. W and X Y and Z
Two protons move parallel to each other with an equal velocity `v=300kms^-1`. Find the ratio of forces of magnetic and electric interaction of the protons.
Figure. Shows a small loop carrying a current I. The curved portion is an arc of a circle of radius R and the straight portion is a chord to the same circle subtending an angle `theta`. The magnetic induction at centor O is A. zeroB. always inward irrespective of the value `theta`C. inward as long as `theta` is less than `pi`D. always outward irrespective of the value of `theta`
A cylindrical cavity of diameter a exists inside a cylinder of diameter `2a` as shown in the figure. Both the cylinder and the cavity are infinitity long. A uniform current density `j` flows along the length . If the magnitude of the magnetic field at the point `P` is given by `(N)/(12) mu_(0)aJ`, then the value of `N` is

Find magnetic field at point P shown in Fig. point P is on the bisector of angle between the wires.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment