A uniform electric field and uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

will turn towards right of direction of motion
speed will decrease
speed will increase
will turn towards left direction of motion

Solution

\(\overrightarrow{V}\) and \(\overrightarrow{B}\) are in same direction so that magnetic force on electron becomes zero,only electric force acts. But force on electron due to electric field is opposite to the direction of velocity.

A current carrying loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, what is the force on the arm AC?

\(-\sqrt{2}\overrightarrow{F}\)
\(-\overrightarrow{F}\)
\(\overrightarrow{F}\)
\(\sqrt{2}\overrightarrow{F}\)

Solution

A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is i., the resultant magnetic field due to the two semicircular parts at their common centre is

\(\frac{\mu _{0}i}{\sqrt{2}R}\)
\(\frac{\mu _{0}i}{2\sqrt{2}R}\)
\(\frac{\mu _{0}i}{2R}\)
\(\frac{\mu _{0}i}{4R}\)

Solution

A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is \(\overrightarrow{F}\), the net force on the remaining three arms of the loop is

\(3\overrightarrow{F}\)
\(-\overrightarrow{F}\)
\(-3\overrightarrow{F}\)
\(\overrightarrow{F}\)

Solution

A very long straight wire carries a current I. At the instant when a charge + Q at point P has velocity \(\overrightarrow{V}\), as shown,the force on the charge is

along oy
opposite to oy
along ox
opposite to ox

Solution

A coil of one turn is made of a wire of certain length and then from the same length a coil of two turns is made.If the same current is passed in both the cases, then the ratio of the magnetic inductions at their centres will be

2 : 1
1 : 4
4 : 1
1 : 2

Solution

A 10eV electron is circulating in a plane at right angles to a uniform field at a magnetic induction 10–4Wb/m²(= 1.0 gauss). The orbital radius of the electron is

12 cm
16 cm
11 cm
18 cm

Solution

[Hint ⇒ ^mv²⁄_r = qvB].

The magnetic field (dB) due to small element (dl) at a distance (\(\overrightarrow{r}\)) from element carrying current i,is

Two equal electric currents are flowing perpendicular to each other as shown in figure. AB and CD are perpendicular to each other and symmetrically placed with respect to the currents. Where do we expect the resultant magnetic field to be zero?

on AB
on CD
on both AB & CD
on both OD & BO

Solution

The direction of the magnetic field due to a current is given by right hand curled fingers rule. Therefore at AB axis, the components of magnetic field will cancel each other and the resultant magnetic field will be zero on AB.

A straight wire of diameter 0.5 mm, carrying a current of 1A is replaced by another wire of 1mm diameter carrying the same current. The strength of magnetic field far away is

unchanged
quarter of its earlier value
half of the earlier value
twice the earlier value

Solution

[Hint ⇒ B ⇒ \(\frac{\mu _{0}i}{2\pi r}\), wherer is distance of point from the wire,where we want to calculate the magnetic field. It is clear from expression that B is independent of thicknessof wire.]

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