Tag: motional emf

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Questions Related to motional emf

Multiple choice motional emf physics

 A coil of circular cross-section having $100$ turns and $4 \mathrm { cm } ^ { 2 }$ face area is placed with its axis parallel to a magnetic field which decreases by $10 ^ { - 2 } \mathrm { Wb } \mathrm { m } ^ { - 2 }$ in $0.01 \mathrm { s }$. The e.m.f induced in the coil is: 

  1. $200 \mathrm { mV }$
  2. $0.4 \mathrm { mV }$
  3. $400 \mathrm { mV }$
  4. $4 \mathrm { mV }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice motional emf physics

Radius of current carrying coil is 'R'. The ratio of magnetic field at a axial point which is R distance away from the centre of the coil to the magnetic field at the centre of the coil :-

  1. $\lgroup \frac{1}{2} \rgroup^{1/2}$
  2. $\frac{1}{2}$
  3. $\lgroup \frac{1}{2} \rgroup^{3/2}$
  4. $\frac{1}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} \vec { B } \, at\, R\, dis\tan  ce=\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2{ { \left( { { x^{ 2 } }+{ R^{ 2 } } } \right)  }^{ 3/2 } } } }  \ =\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2{ { \left( { 2{ R^{ 2 } } } \right)  }^{ 3/2 } } } }  \ =\frac { { { \mu _{ o } }I{ R^{ 2 } } } }{ { 2.{ { \left( 8 \right)  }^{ 1/2 } }.{ R^{ 3 } } } }  \ =\frac { { { \mu _{ o } }i } }{ { 4\sqrt { 2 } R } }  \ \vec { B } \, \, at\, \, centre=\frac { { { \mu _{ o } }i } }{ { 2R } }  \ Ratio=\frac { { { \mu _{ o } }i.2R } }{ { 4\sqrt { 2 } R.{ \mu _{ o } }i } } =\frac { 1 }{ { 2\sqrt { 2 }  } } ={ \left( { \frac { 1 }{ 2 }  } \right) ^{ 3/2 } } \ Hence, \ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

Multiple choice motional emf physics

The magnetic induction due to a magnet on the equatorial line at a distance 0.2 m is $54 \times 10^{-6}$T. The magnetic induction at 0.3m is

  1. $1.6 \times 10^{-6} \quad T$
  2. $1.6 \times 10^{-5} \quad T$
  3. $3.2 \times 10^{-6} \quad T$
  4. $3.2 \times 10^{-5} \quad T$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Magnetic induction on the equatorial line is proportional to 1/r^3. B1 / B2 = (r2 / r1)^3. (54 * 10^-6) / B2 = (0.3 / 0.2)^3 = (1.5)^3 = 3.375. B2 = (54 * 10^-6) / 3.375 = 16 * 10^-6 = 1.6 * 10^-5 T.

Multiple choice motional emf physics

A uniform metal rod is moving with a uniform velocity $v$ parallel to a long straight wire carrying a current $I$. The rod is perpendicular to the wire with its ends at distances $r _{1}$ and $r _{2}$ with $(r _{2} > r _{1})$ from it. The emf induced in the rod is

  1. Zero
  2. $\dfrac {\mu _{0}Iv}{2\pi}\log _{e}\left (\dfrac {r _{2}}{r _{1}}\right )$
  3. $\dfrac {\mu _{0}Iv}{2\pi}\log _{e}\left (\dfrac {r _{1}}{r _{2}}\right )$
  4. $\dfrac {\mu _{0}Iv}{4\pi}\left (1 - \dfrac {r _{1}}{r _{2}}\right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The induced emf in a rod moving in a magnetic field is given by the integral of (v x B) dot dl. For a wire carrying current I, the magnetic field at distance r is B = (mu_0 * I) / (2 * pi * r). Integrating this from r1 to r2 yields the formula in option B.

Multiple choice motional emf physics

If the permeability of iron piece is $3 \times 10 ^ { - 3 }$ and intensity of magnetising field of iron piece is 120 A/m, then what is the magnetic induction of iron piece 

  1. $0.36 \mathrm { Wb } / \mathrm { m } ^ { 2 }$
  2. $5 \times 10 ^ { - 3 } \mathrm { Wb } / \mathrm { m } ^ { 2 }$
  3. $40 \mathrm { Wb } / \mathrm { m } ^ { 2 }$
  4. $2.5 \times 10 ^ { - 4 } \mathrm { Wb } / \mathrm { m } ^ { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Magnetic induction B is given by B = mu * H, where mu is the permeability and H is the intensity of the magnetising field. Given mu = 3 * 10^-3 and H = 120 A/m, B = (3 * 10^-3) * 120 = 0.36 Wb/m^2.

Multiple choice motional emf physics

A conducting ring of radius r is placed perpendicularly inside a time verying magnetic field given by $B={ B } _{ 0 }+\alpha t.{ B } _{ 0 }$ and $\alpha $ are positive constants. E. m. f induced in the ring is 

  1. $-\pi \alpha r$
  2. $-\pi \alpha { r }^{ 2 }$
  3. $-\pi { \alpha }^{ 2 }{ r }^{ 2 }$
  4. $-\pi { \alpha }^{ 2 }{ r }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Faraday's law states emf = -d(phi)/dt. The magnetic flux phi = B * A = (B_0 + alpha * t) * (pi * r^2). Differentiating with respect to time gives emf = -pi * r^2 * alpha.

Multiple choice motional emf physics

In the figure magnetic points into the plane of paper and the rod of length $l$ is moving in the field such that the bottom most point has a velocity $v _1$ and the topmost point has the velocity $V _2(V _2>V _1)$ The emf induced is given by 

  1. $Bv _1l$
  2. $Bv _2l$
  3. $\cfrac 1 2 B(v _2+v _1)l$
  4. $\cfrac 1 2 B(v _2-v _1)l$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Velocity of a point at a distance $x$ from the bottom is given by,

$v=v _1+(\dfrac{v _1-v _2}{l})x$

Potential difference on a small length at this distance is $e=Bx(dx)$

Therefore,

Total potential difference, $e=\int^l _0 Bvdx$

$e=\int^l _0 B(v _1+\dfrac{(v _2-v _1)x}{l})dx$

$=\dfrac{B(v _1+v _2)l}{2}$
Multiple choice motional emf physics

A uniform magnetic field exists in region given by $\vec B = 3\hat i + 4\hat j + 5\hat k$. A rod of length $5m$ is placed along $y$ moved along $x-axis$ with constant speed $1m/sec$. Then induced e.m.f. in the rod will be:

  1. $zero$
  2. $25V$
  3. $20V$
  4. $15V$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The induced emf is given by E = (v x B) dot L. Here, v = 1 * i (m/s), B = 3i + 4j + 5k (T), and L = 5 * j (m). The cross product v x B = (1 * i) x (3i + 4j + 5k) = 4k - 5j. Then E = (4k - 5j) dot (5j) = -25V. The magnitude is 25V.

Multiple choice motional emf physics

A square coil of side 0.5$\mathrm { m }$ has movable side It is placed such that its plane is perpendicularuniform magnetic field of induction 0.2$\mathrm { T }$ . If all sides are allowed to move with a speed of 0.1for 4 sec outwards, average indaced emf is

  1. Zero
  2. 0.01$\mathrm { V }$
  3. 0.028$\mathrm { V }$
  4. 0.072$\mathrm { V }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The change in flux is Delta(phi) = B * Delta(Area). The side increases from 0.5m to 0.5 + (0.1 * 4) = 0.9m. The area changes from 0.5^2 = 0.25 to 0.9^2 = 0.81. Delta(phi) = 0.2 * (0.81 - 0.25) = 0.2 * 0.56 = 0.112. Emf = Delta(phi) / Delta(t) = 0.112 / 4 = 0.028V.

Multiple choice motional emf physics

The amplitude of a magnetic field, which is part of a harmonic electromagnetic wave in vacuum, is  $\mathrm { B } _ { 0 } =510\mathrm { nT } .$  What is the amplitude of the electric field of the wave? 

  1. $140 \mathrm { NC } ^ { - 1 }$
  2. $153 \mathrm { NC } ^ { - 1 }$
  3. $163 \mathrm { NC } ^ { - 1 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an electromagnetic wave, the relationship between electric field amplitude E_0 and magnetic field amplitude B_0 is E_0 = c * B_0, where c is the speed of light (3 * 10^8 m/s). E_0 = (3 * 10^8) * (510 * 10^-9) = 153 N/C.