A charged particle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic moment µis given by

qvR²
qvR²/2
qvR
qvR/2

Solution

Four wires, each of length 2.0 m, are bent into four loops P,Q, R and S and then suspended in a uniform magnetic field.If the same current is passed in each, then the torque will be maximum on the loop

Solution

For a given perimeter the area of circle is maximum. So magnetic moment of (S) is greatest.

A wire ABCDEF is bent in the form as shown in figure. The wire carries a current I and is placed in a uniform magnetic field of induction B parallel to positive Y-axis.If each side is of length L, the force experienced by the wire will be

IBL along the positive Z-direction
IBL along the negative Z-direction
2IBL along the positiveZ-direction
2IBL along the negative Z-direction

An electron traveling with a speed u along the positive x-axis enters into a region of magnetic field where B = –\(\hat{K}\)( x > 0). It comes out of the region with speed v then

v = u at y > 0
v = u at y < 0
v > u at y > 0
v > u aty < 0

Solution

A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field \(\underset{B_{0}}{\rightarrow}\) such that \(\underset{B_{0}}{\rightarrow}\) is perpendicular to the plane of the loop. The magnetic force acting on the loop is

ir B₀
2π ir B₀
zero
π ir B₀

Solution

The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Fleming's Left hand rule on element AB the direction of force will be Leftwards and the magnitude will be

dF = Idl B sin 90° = IdlB

On element CD, the direction of force will be towards right on the plane of the papper and the magnitude will be dF = IdlB.

Three wires are situated at the same distance. Acurrent of 1A, 2A, 3A flows through these wires in the same direction.What is ratio of F₁/F₂, where F₁ is force on 1 and F₂ on 2?

7/8
1
9/8
None of these

Solution

Due to flow of current in same direction at adjacent side, an attractive magnetic force will be produced.

A long wire is bent into shape ABCDE as shown in fig., with BCD being a semicircle with centre O and radiusr metre. A current of I amp. flows through it in the direction A →B →C →D →E. Then the magnetic induction at the point O of the figure in vacuum is

\(\mu _{0}\left [ \frac{1}{2\pi r}+\frac{1}{4r} \right ]\)
\(\mu _{0}\left [ \frac{1}{2\pi r}-\frac{1}{4r} \right ]\)
μ₀I/4r
μ₀I/r

Solution

An infinite straight conductor carrying current 2 I is split into a loop of radius r as shown in fig. The magnetic field at the centre of the coil is

\(\frac{\mu _{0}}{4\pi }\frac{2(\pi +1)}{r}\)
\(\frac{\mu _{0}}{4\pi }\frac{2(\pi -1)}{r}\)
\(\frac{\mu _{0}}{4\pi }\frac{2(\pi -1)}{r}\)
zero

Solution

Here, the wire does not produce any magnetic field at O because the conductor lies on the line of O.Also, the loop does not produce magnetic field at O.

The field B at the centre of a circular coil of radius r istimes that due to a long straight wire at a distance r from it,for equal currents. Fig. shows three cases:

in all cases the circular part has radius r and straight ones are infinitely long. For same current the field B at the centre P in cases 1, 2, 3 has the ratio

\(\left ( -\frac{\pi }{2} \right ):\frac{\pi }{2} :\left ( \frac{3 \pi }{4}- \frac{1}{2} \right )\)
\(\left ( -\frac{\pi }{2}+1 \right ):\left ( \frac{\pi }{2}+1 \right ):\left ( \frac{3 \pi }{4}+ \frac{1}{2} \right )\)
\(\left ( -\frac{\pi }{2}+1 \right ):\left ( \frac{\pi }{2}+1 \right ):\left ( \frac{3 \pi }{4}+ \frac{1}{2} \right )\)
\(\left ( -\frac{\pi }{2}-1 \right ):\left ( \frac{\pi }{4}+\frac{1}{4} \right ):\left ( \frac{3 \pi }{4}+ \frac{1}{2} \right )\)

Solution

In fig, what is the magnetic field induction at point O

\(\frac{\mu _{0}i}{4\pi r}\)
\(\frac{\mu _{0}i}{4\pi r}+\frac{\mu _{0}i}{2\pi r}\)
\(\frac{\mu _{0}i}{4\pi r}+\frac{\mu _{0}i}{4\pi r}\)
\(\frac{\mu _{0}i}{4\pi r}-\frac{\mu _{0}i}{4\pi r}\)

Solution

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