Q: A conducting square loop initially lies in the X Z plane with its lower edge hinged along the X-axis. Only in the region y 0, there is a time dependent magnetic field pointing along the Z-direction, B

aligned & = B _0 tA t = B _0 ~A 2 t 2 \\ & =- d dt =- B _0 ~A 2 t = (0 t ) \\ & =0 ( t 2 ) aligned Ans. Option (A)

Q: A thin conducting rod M N of mass 20 gm , length 25 cm and resistance 10 Ω is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is

Induced emf ε = B l v ⇒ Induced current i = ε R = B l v R Direction of induced current would be from N to M . Now magnetic force acting on the rod due to this current would be in the upward direction. Therefore, we can write ⇒ m g - i l B = m a [Applying 2 nd law] ⇒ m g - B 2 l 2 v R = m d v d t ⇒ ∫ 0 v d v m g - B 2 l 2 v R = ∫ 0 t d t m ⇒ ln m g - B 2 l 2 v R

Q: A long straight wire carries a current, I = 2 ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are paral

Magnetic field due the long wire, B = μ 0 I 2 π x Motional emf in small element, d ε = B v d x = 3 μ 0 2 2 π x d x ⇒ ε = 3 μ 0 π ∫ 1 4 1 x d x ∴ ε = 3 μ 0 π ln 4 = 16.8 × 10 − 7 Maximum charge on capacitor, q max = ε C = 16.8 × 10 − 7 5 × 10 − 6 = 84 × 10 − 13 C = 8.4 × 10 − 12 C Maximum current in circuit, I max

Question 1

List-I contains four conducting loops lying in the XY plane, as shown in the figures. The loops are rotating about Z axis passing through the point O with time period T in clockwise direction. The reg

Accepted Answer

Let us analyze the induced current i(t) for each loop as it rotates clockwise with time period T. The magnetic field B is uniform in the +z direction for x > 0. The angular velocity is = 2 T . The induced EMF is = - d dt = -B dA dt , where A(t) is the area of the loop in the region x > 0. The current i(t) is proportional to dA dt . Loop (P): This is a semicircle in the region x 0. The area in x > 0 is a sector of angle = t. For 0 t T/2, the area A(t) = 1 2 R^2 t increases linearly, so dA dt is a

Question 2

A conducting square loop initially lies in the X Z plane with its lower edge hinged along the X-axis. Only in the region y 0, there is a time dependent magnetic field pointing along the Z-direction, B

Accepted Answer

aligned & = B _0 tA t = B _0 ~A 2 t 2 \ & =- d dt =- B _0 ~A 2 t = (0 t ) \ & =0 ( t 2 ) aligned Ans. Option (A)

Question 3

A conducting square loop of side L, mass M and resistance R is moving in the X Y plane with its edges parallel to the X and Y axes. The region y 0 has a uniform magnetic field, B =B_0 k. The magnetic

Accepted Answer

- d dt = d dt ( B _0 y )= BV aligned & F = B ( i )( )(- j ) \ & ma =- B _0 [ ~B _0 ~V R ]( ) \ & a =- B _0^2 ^2 ~V mR aligned aligned & Also K= B_0^2 ^2 V R M \ & So [a=-k v] \ & d v d t =-k v \ & _ v_0 ^v d v d t = _0^t-k d t \ & n v v_0 =-k t aligned [ v = v _0 e ^ - kt ] [ v = v _0 e ^ - kt ] .( i ) dx dt = v _0 e ^ - kt ( x ) aligned & _0^ x dx = _0^ t v _0 e ^ - kt dt \ & = v _0 k (1- e ^ - kt ) aligned When x= = V _0 k (1- e ^ - kt ) Option (D) ( v _0=3 k ) aligned & = 3 k k (1- e ^

Question 4

A region in the form of an equilateral triangle (in x-y plane) of height L has a uniform magnetic field B pointing in the +z-direction. A conducting loop PQR , in the form of an equilateral triangle o

Accepted Answer

For, 0 to L = B _ eff v = B x 3 v For, L to 2 ~L aligned & |e m f|=B ( L 3 - x_0 3 ) v-B 2 x_0 3 v \ & = B v L 3 - 3 Bvx _0 \ & = Bv [ L 3 - 3 (x-L) ] \ & = B v 3 [L-3 x +3 ~L ] \ & = B v 3 [4 L-3 x ] \ & at x= 4 L 3 \ & e m f=0 \ & aligned

Question 5

A rectangular conducting loop of length 4 &#160; cm and width 2 &#160; cm is in the x y -plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction&

Accepted Answer

The two sides of the rectangular loop which are perpendicular to the wire would not contribute in induced emf. Angle of velocity vector with x-axis is  tan &#945; = 1 2 3 2 = 1 3 . Therefore,  &#945; = 30 &#176; . For parallel sides: E &#8594; = B &#8594; &#215; v &#8594; = &#956; 0 I 2 &#960; x &#215; v &#8658;  Net emf  = E 1 cos 60 o - E 2 cos 60 o &#215; width = 1 2 &#215; 2 100 &#215; &#956; 0 I v 2 &#960; 1 4 100 - 1 8 100 = 1 100 &#215; 2 &#215; 10 - 7 &#215; 10 &

Question 6

A thin conducting rod M N of mass 20 &#160; gm , length 25 &#160; cm and resistance 10 &#937;  is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is

Accepted Answer

Induced emf &#949; = B l v &#8658;  Induced current  i = &#949; R = B l v R Direction of induced current would be from  N &#160; to &#160; M . Now magnetic force acting on the rod due to this current would be in the upward direction. Therefore, we can write &#8658; &#160; m g - i l B = m a  [Applying 2 nd  law] &#8658; &#160; m g - B 2 l 2 v R = m d v d t &#8658; &#8747; 0 v &#160; d v m g - B 2 l 2 v R = &#8747; 0 t d t m &#160; &#8658; &#160; ln m g - B 2 l 2 v R

Question 7

Paragraph A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a co

Accepted Answer

For a super conducting loop, net flux passing through it will remain constant. &#981; total &#160; i = &#981; total &#160; f As area is same, therefore, &#956; 0 m 2 &#960;r 3 = &#956; 0 I 2 a &#8658; I = a m &#960;r 3 &#8658; I &#8733; m r 3

Question 8

A long straight wire carries a current,  I = 2  ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are paral

Accepted Answer

Magnetic field due the long wire,  B = &#956; 0 I 2 &#960; x Motional emf in small element,  d &#949; = B v d x = 3 &#956; 0 2 2 &#960; x d x &#8658; &#949; = 3 &#956; 0 &#960; &#8747; 1 4 1 x d x &#8756; &#160; &#160; &#949; = 3 &#956; 0 &#960; ln 4 = 16.8 &#215; 10 &#8722; 7 Maximum charge on capacitor,  q max = &#949; C = 16.8 &#215; 10 &#8722; 7 5 &#215; 10 &#8722; 6 = 84 &#215; 10 &#8722; 13 &#160; C = 8.4 &#215; 10 &#8722; 12 &#160; C Maximum current in circuit,  I max

Question 9

The inductors of two L R circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are spe

Accepted Answer

ε 1 = L 1 d i 1 d t + M d i 2 d t d W 1 = ε 1 i 1 d t = L 1 d i 1 i 1 + M d i 2 i 1 d W 2 = L 2 d i 2 i 2 + M d i 1 i 2 d W 1 + d W 2 = L 1 i 1 d i 1 + L 2 i 2 d i 2 + M d i 1 · i 2 W = 1 2 L 1 i 1 2 + 1 2 L 2 i 2 2 + M 1 i 2 = 55 mJ , Note : If mutual induction is taken in opposite direction then W = 35 mJ

Question 10

A light disc made of aluminum (a nonmagnetic material) is kept horizontally and is free to rotate about its axis as shown in the figure. A strong magnet is held vertically at a point above the disc aw

Accepted Answer

When we start rotating the magnet above the disc, eddy currents will be generated in the disc. Now we know from Lenz's law that the induced emf will drive the current such that it opposes the cause of the production of induced emf. Hence the disc will rotate in the same direction as the direction of magnet's motion

Electromagnetic Induction — Previous Year Questions & Solutions