Question 1

The straight wire A B carries a current I. The ends of the wire subtend angles _1 and _2 at the point P as shown in figure. The magnetic field at the point P is

Accepted Answer

The magnetic field at a point P due to a current carrying straight wire, such that the ends of the wires, makes angle and (see figure) is given by B= _0 I 4 d ( - ) where, d is the distance between the point P and wire. In the given problem, = _1 and = _2 B= _0 I 4 d ( _1- _2 )

Question 2

A long straight wire of radius a carries a steady current 1 . The current is uniformly distributed across its cross section. The ratio of the magnetic field at a 2 and 2a from axis of the wire is:

Accepted Answer

Magnetic field due to straight wire, B= _0 I 2 r Magnetic field at a 2 is, B _ a / 2 = _0 I 2 ( a / 2) B _ a / 2 ~B _ 2 a = 1 1 =1: 1

Question 3

The electrostatic force ( F _1 ) and magnetic force ( F _2 ) acting on a charge q moving with velocity v can be written:

Accepted Answer

Electrostatic force, F _1= q E Magnetic force, F _2= q ( V B )

Question 4

The magnetic moments associated with two closely wound circular coils A and B of radius r_A=10 ~cm and r_B=20 ~cm respectively are equal if: (Where N _A, I _A and N _ B , I _ B are number of turn and

Accepted Answer

The magnetic moment associated with circular coil, aligned & M=N I A \ & M_A=M_B \ & N_A I_A A_A=N_B I_B A_B \ & N_A I_A (0.1)^2=N_B I_B (0.2)^2 ( A= r^2 ) \ & N_A I_A=4 N_B I_B aligned

Question 5

A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 ~m long and 0.2 ~m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through

Accepted Answer

Given that number of turns, n=1200 Current, I =2 ~A Magnetic field at centre inside the solenoid is given by, B = _0 nI So magnetic intensity at centre of the solenoid, aligned & H = B _0 = nI = ( 1200 2 )(2) ( B = _0 H ) \ & H =1.2 10^3 Am ^ -1 aligned

Question 6

A charge particle is moving in a uniform magnetic field (2 i +3 j ) T . If it has an acceleration of ( i -4 j ) m / s ^2, then the value of will be :

Accepted Answer

Given that uniform magnetic field, B =(2 i +3 j ) T Acceleration a =( i -4 j ) m / s ^2 We know that F=q( v B ) m a =q( v B ) Here, a B , so, a B =0 ( i -4 j )(2 i +3 j )=0 2 -12=0 =6

Question 7

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:

Accepted Answer

As M = IA | M |= e 2 R v R ^2 [ I = Q T = e 2 R V ] | M |= 1 2 e v R | M |= mvR 1 e 2 ~m | M |= eL 2 ~m | M |=- e L 2 ~m [ Here M and L will always be opposite]

Question 8

A proton moving with a velocity 3 10^ 5 ~m / s enters a magnetic field of 0.3 tesla at an angle of 30^ with the field. The radius of curvature of its path will be ( e / m for proton =10^ 8 C / kg )

Accepted Answer

r= m v B e = 3 10^ 5 30^ 0.3 10^ 8 3 10^ 5 1 2 3 10^ 7 =0.5 10^ -2 m=0.5 ~cm

Question 9

A solenoid is 2 ~m long and 3 ~cm in diameter. It has 5 layers of winding of 1000 turns each and carries a current of 5 ~A . What is the magnetic field at its centre ? Use the standard value of _ 0 .

Accepted Answer

Length of solenoid, l=2 ~m Total number of turns, N =5 1000=5000 Number of turns per unit length, n = N l = 5000 2 =2500 ~m ^ -1 Current, I = 5A Now, B = _ 0 nI =4 10^ -7 2500 5 ~T =1.57 10^ -2 ~T

Magnetic Effects of Current — Previous Year Questions & Solutions