In the figure shown, a uniform magnetic field is perpendicular to the plane of circuit. The sliding rod of length l = 0.25m moves …..

Q. In the figure shown, a uniform magnetic field is perpendicular to the plane of circuit. The sliding rod of length l = 0.25m moves uniformly with constant speed v = 4ms-1. If the resistance of the slides is 2Ω, then the current flowing through the sliding rod is

Current Electricity

(a) 0.1A

(b) 0.17A

(c) 0.08A

(d) 0.03A

Ans: (a)

Electric charge q is distributed uniformly over a rod of length l. The rod is placed parallel to a long wire carrying a current i….

Q. Electric charge q is distributed uniformly over a rod of length l. The rod is placed parallel to a long wire carrying a current i. The separation between the rod and the wire is a. The force needed to move the rod along its length with a uniform velocity v is

(a) \displaystyle \frac{\mu_0 i q v}{2 \pi a}

(b) (a) \displaystyle \frac{\mu_0 i q v}{4 \pi a}

(c) (a) \displaystyle \frac{\mu_0 i q v l}{2 \pi a}

(d) (a) \displaystyle \frac{\mu_0 i q v l}{4 \pi a}

Ans: (a)

An LR circuit is connected to a battery at time t = 0. The energy stored in the inductor reaches half its maximum value at time

Q. An LR circuit is connected to a battery at time t = 0. The energy stored in the inductor reaches half its maximum value at time

(a) \displaystyle \frac{R}{L} ln \frac{\sqrt2}{\sqrt2 -1}

(b) \displaystyle \frac{L}{R} ln \frac{\sqrt2-1}{\sqrt2 }

(c) \displaystyle \frac{L}{R} ln \frac{\sqrt2}{\sqrt2 -1 }

(d) \displaystyle \frac{R}{L} ln \frac{\sqrt2-1}{\sqrt2 }

Ans: (c)

A conducting straight wire PQ of length l is fixed along a diameter of a non-conducting ring as shown in the figure…………….

Q. A conducting straight wire PQ of length l is fixed along a diameter of a non-conducting ring as shown in the figure. The ring is given a pure rolling motion on a horizontal surface such that its centre of mass has a velocity v. There exists a uniform horizontal magnetic field B in horizontal direction perpendicular to the plane of ring. The magnitude of induced emf in the wire PQ at the position shown in the figure will be

 Numerical

(a) Bvl

(b) 2Bvl

(c)3Bvl/2

(d) zero

Ans: (a)

A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. Some how the sides of …..

Q. A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. Some how the sides of the loop start shrinking at a constant rate α  . The induced emf in the loop at an instant when its side is a, is

(a) 2a α B

(b) a² α B

(c) 2a² α B

(d) a α B

Ans:(a)