How To Find The Magnetic Field At Point P. Where Two Ellipse Are There And Distance Is D

What is the magnitude of magnetic field at the centre of a hexagonal loop of a metallic wire having each side 'a' and carrying current I (μI√3/Πa)?

Suppose that the current I flows through the hexagonal loop ABCDEF of each side= a.The magnetic field produced at the point O due to each of the six sides of the hexagonal loop is same and perpendicular to its plane (in outward direction).Here is the solution:—If perpendicular distance of the point O from the side AB (that is OM=r), then magnetic field at the point O due to the side AB of the loop= B(AB)The magnetic field due to the whole hexagonal loop, B (loop)= 6 x B(AB)B (loop) = √3μI/πaTushar Raj

How will the magnetic field at the centre of a circular coil carrying a current change if the current is doubled and the radius of the coil is halved?

At the center of a current carrying coil, the magnetic field intensity is directly proportional to the current and inversely proportional to the radius of the coil.So, if the current is doubled and radius is halved, then the value of B increases by 4 times the initial value.

What is the direction of the magnetic field due to a circular current carrying coil?

To find out the direction of a magnetic field in a coil carrying current, let's use a step by step process starting from the field produced by a straight wire carrying current.Magnetic field of a wireA current carrying wire produces circular lines of force centered at the wire. The direction of the current is by convention the direction in which positive charges move.The direction of the magnetic field is in the direction the fingers of your right hand would curl, if you wrapped them around the wire with your thumb in the direction of the current.Magnetic field of a loopIf we bend the straight wire to form a loop, the circular lines of force of the wire combine to form lines more similar to the ones of a bar magnet.We can still use the right hand rule to determine the direction of the magnetic field.Magnetic field of a coilWe can increase the field strength by winding multiple turns of wire into a coil.The direction of the field is the same as the one of a single turn with the current flowing in the same way.We can directly find the direction of the magnetic field by using another right hand rule. If we curl our fingers in the direction of the current, our thumb indicates the direction of the magnetic field.

L1 and L2 are mutually fixed perpendicular lines. P is a variable point in the the plane such that the sum distances of P from L1 and L2 is 3. What is the area enclosed by the locus P?

The locus of P is by definition is that of the ellipse. If we take L1 and L2 as x-axis and y-axis respectively, then P will describe an ellipse.Hence the area enclosed by locus of P is that of the ellipse which is pi (ab).

In electric dipole moment P=q*d, how can I calculate q if I have two charges, one positive and one negative?

That’s only the electric dipole moment for equal charges. In general, the expression is[math] \vec p = \sum_i q_i \vec r_i [/math]So if you have charges [math]q_1 [/math] and [math] q_2 [/math] located at positions [math] \vec r_1 = \frac{d}{2} \hat z [/math] and [math] \vec r_2 = -\frac{d}{2} \hat z [/math], then your dipole moment will be[math] \vec p = \frac{(q_1 - q_2) d}{2} \hat z[/math]and its magnitude will be[math]p = \frac{|q_1 - q_2| d}{2} [/math]which reduces back to [math] p = qd [/math] in the case that the charges are equal and opposite.

What is the formula for electric dipole?

1.Electric dipole moment:-p=qa, where p is electric dipole moment, q=one of the charge of the dipole and a=distance between the two charges.2.Electric field due to a dipole at any point:E = 1/4πε0 P/r^3(3Cos^2θ + 1)^1/2. where, p=electric dipole moment, r=distance of the point from the center of the dipole and θ=angle subtended by the distance vector at the dipole.3.Potential due to a dipole at any point:V=pcosθ/4πε0(r^2 - a^2cos^2θ).

A particle of mass m and charge q is revolving around a fixed charge -Q in a circular path of radius R. What will be the frequency of revolution?

So it is given that ,There is a negative charge [math]Q[/math] is at the centre of the circle. A charge of [math]q[/math] having mass m is revolving around this charge centre [math] -Q[/math].As the centre has a negative charge so the force will be inward.Electric force will be,[math]\displaystyle F_{e}=\frac{KQq}{R^2}[/math]Also as the charge has mass [math]m[/math] and it is revolving around in a circular path so there will be centripetal force acting too,whose magnitude is,[math]\displaystyle F_{c}=\frac{mv^2}{R}[/math]According to the question,[math]\displaystyle F_{e}=F_{c}\tag*{}[/math][math]\displaystyle \Rightarrow \frac{KQq}{R^2}=\frac{mv^2}{R}[/math]From this we can get,[math]v=\displaystyle \sqrt{\frac{KQq}{mR}}[/math]The charge [math]q[/math] revolves around the circle ,so distance covers in one rotation is [math]2\pi R.[/math]We have the velocity and the distance covered. We can find the time period of motion.[math]\displaystyle T=\displaystyle \frac{s}{v}=\displaystyle \frac{2 \pi R}{\displaystyle \sqrt{\frac{KQq}{mR}}}[/math]Now you can put the formula of frequency,[math]f=\displaystyle \frac{1}{T}\tag*{}[/math]We will get,[math]\displaystyle \boxed{f=\displaystyle \frac{\sqrt{KQq}}{2 \pi R \sqrt{mR}}}[/math]That's it.[math]Tripathy.[/math]

The ends of a rod of length l move on two mutually perpendicular lines. How do you find the locus of the point on the rod which divides it in the ratio 1:2?

Let the rod be lying on the X and Y axes. Let the point of contact on X axis be (a,0) and that on the Y axis be (0,b)Let the point on the rod be (h,k)As the point divides the rod in the ratio 1:2, by section formula we have,h=a/3or a=3h. eqn 1k =2b/3or b =3k/2. eqn 2Now by Pythagoras theorem,a^2 + b^2 = L^2Replacing a and b by eqn 1 and eqn 2 respectively, we get,9(h^2) + 9(k^2)/4 = L^2Now replace h,k by x, y to get the locus.