Informative line

Basic Concepts Of Motional Emf

Practice equation to find the potential difference of a circular loop moving in magnetic field, and motional EMF in conducting rod. Learn dependence of magnitude of induced EMF on angle between velocity and length of rod.

Motional EMF

  • Consider a rod of length '\(\ell\)' which  is moving with velocity '\(v\)' in a uniform magnetic field, as shown in figure.

Motional EMF

The  emf generated due to the motion  of rod is known as motional EMF. It is given by, 

\(|\mathcal{E}| = B\,\ell v\)

where,

\(B\)  is magnetic field 

\(\ell\)  is length of rod 

\(v\)  is velocity of rod and

\(\vec B, \vec \ell, \vec v\) are mutually perpendicular

 

Dependence of Magnitude of Induced EMF on Angle between Velocity and Length of Rod

The angle between velocity and length of rod is not 90º.

Thus, the component of velocity is so chosen that it should be perpendicular to the length of rod.

\(|\mathcal{E}| = v_{\bot} B\,\ell = B \,\ell\, v \,sin\, \theta\)

Illustration Questions

A rod of length \(\ell\) is moving with velocity \(v\) in a uniform magnetic field \(B\). Initially the velocity of rod making an angle \(\theta=\) 90º with its length. Now, the velocity of rod makes an angle \(\theta=\)30º with its  length, as shown in figure. What will be the change in magnitude of emf?

A \(\dfrac{B\ell v}{2}\)

B \(\dfrac{B\ell v}{3}\)

C \(\dfrac{B\ell v}{4}\)

D \(\dfrac{B\ell v}{5}\)

×

Length of rod  = \(\ell\)

velocity of rod = \(v\)

image

Magnitude of emf at initial position,

\(\mathcal{E}_i = B\ell v\)

image

Magnitude of emf at final position,

\(\mathcal{E}_f = B \ell v \,sin\theta\)

\(\mathcal{E}_f = B \,\ell \,v \,(sin\,30º)\)

\(\mathcal{E}_f = \dfrac{B\ell v}{2}\)

image

Change in magnitude of emf,

\(\Delta \mathcal{E} = |\mathcal{E}_f - \mathcal{E}_i|\)

\(\Delta \mathcal{E} = \bigg|\dfrac{B\ell v}{2} - B\ell v\bigg|\)

\(\Delta\, \mathcal{E} = \dfrac{B\ell v}{2}\)

image

A rod of length \(\ell\) is moving with velocity \(v\) in a uniform magnetic field \(B\). Initially the velocity of rod making an angle \(\theta=\) 90º with its length. Now, the velocity of rod makes an angle \(\theta=\)30º with its  length, as shown in figure. What will be the change in magnitude of emf?

image
A

\(\dfrac{B\ell v}{2}\)

.

B

\(\dfrac{B\ell v}{3}\)

C

\(\dfrac{B\ell v}{4}\)

D

\(\dfrac{B\ell v}{5}\)

Option A is Correct

Motional EMF of a Side of a Rectangular Loop

  • Consider a rectangular loop PQRS which is moving in uniform magnetic field, as shown in figure.

  • The initial velocity of rectangular loop PQRS is \(v\).
  • Motional EMF of side QR = \(B\ell v\)  as \(\vec B \,\bot\,\vec v\, \bot\,\vec \ell \) 
  • Motional EMF of side RS = 0

          Since, there is no component of velocity which is perpendicular to the length.

          Thus, there is no motional EMF.

  • Motional EMF of side SP = \(B\ell v\)  as  \(\vec B \,\bot\,\vec v\, \bot\,\vec \ell \) 
  • Motional EMF of side PQ = 0  as  \(\vec v \parallel \vec \ell\)

          Since, there is no component of velocity which is perpendicular to the length.

          Thus, there is no motional EMF.

  • If any two of \(\vec B\),\(\vec l\)and \(\vec v\) are zero, then motional emf is zero.

Illustration Questions

A rectangular loop of dimensions \(\ell × b =\) 3 cm × 4 cm is moving with initial velocity \(v=\) 3 cm/sec in a uniform magnetic field region of \(B\) = 1T. Find the motional EMF of side CD, as shown in figure.

A 2 V

B 3 V

C Zero

D 4 V

×

Dimensions of rectangular loop,  \(\ell × b =\) 3 cm × 4 cm

Initial velocity of loop, \(v=\) 3 cm/sec

Magnetic field, \(B\) = 1 T

image

The motional EMF of side CD is zero. 

Since, there is no component of velocity which is perpendicular to the length.

Thus, there is no motional EMF.

 

image

A rectangular loop of dimensions \(\ell × b =\) 3 cm × 4 cm is moving with initial velocity \(v=\) 3 cm/sec in a uniform magnetic field region of \(B\) = 1T. Find the motional EMF of side CD, as shown in figure.

image
A

2 V

.

B

3 V

C

Zero

D

4 V

Option C is Correct

Potential Difference of a Circular Loop moving in Magnetic Field

  • Consider a circular coil of radius r which is moving with velocity \(v\) in uniform magnetic field B, as shown in figure.

Potential difference between point P and Q 

  • The potential difference between the points P and Q is given by,

 \(V _{PQ} = B\, \ell \,v \)         (as  \(\vec B \,\bot\,\vec v\, \bot\,\vec \ell \))

          = \(B × 2R× v\)

          = \(2 \,B \, R \, v \)

Illustration Questions

A circular coil of radius r is moving with velocity \(v\) into a region of uniform magnetic field B, as shown in figure. Find the potential difference between S and P.

A \(B\,R\,v\)

B \(2\,B\,R\,v\)

C \(\dfrac{B R v}{2}\)

D 0

×

Here, points S and P are at same potential because \(\vec \ell\) is parallel to  \(\vec v\)

\(\Delta V _{SP} = 0\)

image image

Option D is Correct.

image

A circular coil of radius r is moving with velocity \(v\) into a region of uniform magnetic field B, as shown in figure. Find the potential difference between S and P.

image
A

\(B\,R\,v\)

.

B

\(2\,B\,R\,v\)

C

\(\dfrac{B R v}{2}\)

D

0

Option D is Correct

Induced EMF of Small Element

  • Consider a conducting rod of length \(\ell\) which rotates with a constant angular speed \(\omega\) about a pivot at one end.
  • A uniform magnetic field \(\vec B\)  is directed perpendicularly to the plane of rotation , as shown in figure.
  • Consider a small element PQ of rod, which is at \(x\) distance from O. PQ  is also moving  with same velocity. Hence, it will generate induced emf too. 

 

Induced EMF by small Element PQ 

 \(\mathcal{E} _{PQ} = B\,v \,dx\)               

[\(\because\) length of PQ is \(dx\)]

The  Relation between Linear  Velocity and Angular Velocity

\(v = x \,\omega\)

Thus , \(\mathcal{E} _{PQ} = B\,x \,\omega\, dx\)

 EMF of Rod

 The sum of induced emf of all small elements will be equal to the emf of rod.

 Thus, \(\mathcal{E}_ {rod} = \displaystyle\int\limits^\ell_0 B \,x \,\omega \, dx\)

 \(\mathcal{E} _{rod} = B\omega \left[\dfrac{x^2}{2}\right]^\ell_0\)

 \(\mathcal{E} _{rod} = B \,\omega \dfrac{\ell^2}{2}\)

 \(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \)

 

Illustration Questions

A rod of length \(\ell = \) 5 m rotates with a constant angular speed \(\omega =\) 2 rad/sec about a pivot at one end in a uniform magnetic field of \(B\) = 1 T, which is directed perpendicularly inwards to the plane of rotation. Find the induced emf of small element PQ of length \(dx=\) 2 cm on the rod whose distance from O is \(x = \) 4 m, as shown in figure.

A \(0.30\, V\)

B \(0.20\, V\)

C \(0.25 \,V\)

D \(0.16 \,V\)

×

Length of rod , \(\ell = \) 5 m 

Angular speed of rod , \(\omega =\) 2 rad/sec

Magnetic field , \(B\) = 1 T

Length of small element , \(dx = \) 2 cm

Distance of small element  from O, \(x=\) 4m

image

Induced EMF of small Element

\(\mathcal{E} _{PQ } = Bv × length \,of \, PQ\)

\(\mathcal{E} _{PQ} = B (x\,\omega) × length \, of \, PQ\)

\(\mathcal{E}_{PQ} = 1 × 4× 2× 2× 10^{-2}\)

\(\mathcal{E} _{PQ} = 16× 10^{-2}\)

\(\mathcal{E} _{PQ} = 0.16\,V\)

image

A rod of length \(\ell = \) 5 m rotates with a constant angular speed \(\omega =\) 2 rad/sec about a pivot at one end in a uniform magnetic field of \(B\) = 1 T, which is directed perpendicularly inwards to the plane of rotation. Find the induced emf of small element PQ of length \(dx=\) 2 cm on the rod whose distance from O is \(x = \) 4 m, as shown in figure.

image
A

\(0.30\, V\)

.

B

\(0.20\, V\)

C

\(0.25 \,V\)

D

\(0.16 \,V\)

Option D is Correct

Motional EMF Induced Between the Ends of a Rotating Rod 

  • Consider a conducting rod of length \(\ell\) which rotates with a constant angular speed \(\omega\) about a pivot at one end.
  • A uniform magnetic field \(\vec B\) is directed perpendicularly to the plane of rotation, as shown in figure.
  • Consider a small element PQ of rod, which is at  \(x\) distance from O. PQ  is also moving  with same velocity. Hence, it will generate induced emf too. 

Induced EMF by Small Element PQ

  \(\mathcal{E} _{PQ} = B\,v \,dx\)               

[\(\because\) length of PQ is \(dx\)]

The  Relation between Linear Velocity and Angular Velocity

\(v = x \,\omega\)

Thus, \(\mathcal{E} _{PQ} = B\,x \,\omega\, dx\)

  EMF of Rod

          The sum of induced emf of all small elements will be equal to the emf of rod.

          Thus, \(\mathcal{E}_ {rod} = \displaystyle\int\limits^\ell_0 B \,x \,\omega \, dx\)

                     \(\mathcal{E} _{rod} = B\omega \left[\dfrac{x^2}{2}\right]^\ell_0\)

                     \(\mathcal{E} _{rod} = B \,\omega \dfrac{\ell^2}{2}\)

                      \(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \)

 

Illustration Questions

A conducting rod  of length \(\ell\) rotates with a constant angular speed \(\omega\) about a pivot at one end. A uniform magnetic field  \(\vec B\) is directed perpendicularly to the plane of rotation, as shown in figure. Find the motional emf induced between the ends of rod.

A \(\dfrac{1}{2}\, B\, \omega\, \ell^2\)

B \(\dfrac{1}{3}\, B\, \omega\, \ell^2\)

C \(\dfrac{3}{2}\, B\, \omega\, \ell^2\)

D \(2\, B\, \omega\, \ell^2\)

×

Consider a small element PQ of rod, which is at \(x\) distance from O. PQ is also moving with same velocity. Hence, it will generate induced emf too. 

image

Induced emf by small element PQ  

\(\mathcal{E} _{PQ} = B\,v \,dx\)               

[\(\because\) length of PQ is \(dx\)]

image

The  relation between linear velocity and angular velocity

 \(v = x \,\omega\)

Thus, \(\mathcal{E} _{PQ} = B\,x \,\omega\, dx\)

 

image

The sum of induced emf of all small elements will be equal to the emf of rod.

Thus, \(\mathcal{E}_ {rod} = \displaystyle\int\limits^\ell_0 B \,x \,\omega \, dx\)

\(\mathcal{E} _{rod} = B\omega \left[\dfrac{x^2}{2}\right]^\ell_0\)

\(\mathcal{E} _{rod} = B \,\omega \dfrac{\ell^2}{2}\)

\(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \)

 

image

A conducting rod  of length \(\ell\) rotates with a constant angular speed \(\omega\) about a pivot at one end. A uniform magnetic field  \(\vec B\) is directed perpendicularly to the plane of rotation, as shown in figure. Find the motional emf induced between the ends of rod.

image
A

\(\dfrac{1}{2}\, B\, \omega\, \ell^2\)

.

B

\(\dfrac{1}{3}\, B\, \omega\, \ell^2\)

C

\(\dfrac{3}{2}\, B\, \omega\, \ell^2\)

D

\(2\, B\, \omega\, \ell^2\)

Option A is Correct

Illustration Questions

A metal bar of length \(\ell\) rotates with angular velocity \(\omega\) about a pivot at one end. A uniform magnetic field \(\vec B\) is perpendicular to the plane of rotation. What is the potential difference between P and Q?  

A  \(\dfrac{B \omega}{2} (y^2 + 2\,xy)\)

B \(\dfrac{B \omega^2}{2} (x^2 + y^2)\)

C \(\dfrac{B \omega^2}{2} (x^2 + 2\,xy)\)

D \(B\, \omega^2\, x y\)

×

The electric field generated inside the rod points towards the pivot, so its radial component is 

 \(E_r = - vB\)

\(\because \,\,\,\,v = r \omega\)

\(\therefore \,\,\,E_r = -r \omega\, B\)                 ..........(1)

image

The potential difference between the ends of rod is given by 

VPQ = VP – VQ

VPQ =  \(\displaystyle -\int \limits^{x+y}_x E_r \cdot dr\)

VPQ =   \(\displaystyle -\int \limits^{x+y}_x (-r \,\omega\, B)\, dr\)               from equation (1)

VPQ =   \(\displaystyle B \,\omega\int \limits^{x+y}_x r \,\, dr\)         

VPQ =   \(B \,\omega \left[\dfrac{r^2}{2}\right]_x^{x+y}\)

VPQ =  \(\dfrac{B \,\omega}{2} \left[ (x+y)^2 - x^2\right]\)

VPQ =   \(\dfrac{B \,\omega}{2} \left[ y^2+2xy \right]\)

 

image

A metal bar of length \(\ell\) rotates with angular velocity \(\omega\) about a pivot at one end. A uniform magnetic field \(\vec B\) is perpendicular to the plane of rotation. What is the potential difference between P and Q?  

image
A

 \(\dfrac{B \omega}{2} (y^2 + 2\,xy)\)

.

B

\(\dfrac{B \omega^2}{2} (x^2 + y^2)\)

C

\(\dfrac{B \omega^2}{2} (x^2 + 2\,xy)\)

D

\(B\, \omega^2\, x y\)

Option A is Correct

Induced EMF across Spoke of a Wheel 

  • Consider a circular wheel having some radial metallic spokes.
  • The length of each spoke is \(\ell\).
  • The wheel is rotating with angular velocity \(\omega\) in a region of uniform magnetic field.
  • Each spoke acts like a rod of length \(\ell\).

  • Consider a small element PQ of rod, which is at \(x\) distance from O. PQ is also moving  with same velocity. Hence, it will generate induced emf too. 

Induced EMF by small Element PQ 

\(\mathcal{E} _{PQ} = B\,v \,dx\)             

[\(\because\) length of PQ is \(dx\)]

The  Relation between Linear Velocity and Angular Velocity

\(v = x \,\omega\)

Thus, \(\mathcal{E} _{PQ} = B\,x \,\omega\, dx\)

 EMF of Rod

  • The sum of induced emf of all small elements will be equal to the emf  of rod.

          Thus, \(\mathcal{E}_ {rod} = \displaystyle\int\limits^\ell_0 B \,x \,\omega \, dx\)

                     \(\mathcal{E} _{rod} = B\omega \left[\dfrac{x^2}{2}\right]^\ell_0\)

                     \(\mathcal{E} _{rod} = B \,\omega \dfrac{\ell^2}{2}\)

                      \(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \)

  • Hence, emf across each spoke is \(\mathcal{E} _{rod} = \dfrac{1}{2}\,B \,\omega\,\ell^2 \) 
  • Hence, each spoke behaves as battery of emf \( \dfrac{1}{2}\,B \,\omega\,\ell^2 \)

 

Illustration Questions

A Ferris Wheel  has radial metallic spokes between the hub and the circular rim. These spokes move in the magnetic field of the earth. If the length of each spoke is \(\ell = \) 10 m and angular velocity is \(\omega = \) 1 rad/s, then find the induced emf across center of wheel and periphery. Given : Magnetic field of Earth, \(B _{earth} = 0.5 × 10^{-4 } \,T\)  (Assume that Bearth is horizontal at the location)

A \(5 \,mV\)

B \(2.5 \,mV\)

C \(3 \,mV\)

D \(4 \,mV\)

×

Length of each spoke, \(\ell = \) 10 m

Angular velocity of spoke, \(\omega =\) 1 rad/s

Magnetic field of Earth, \(B_{earth} = 0.5 × 10^{-4} \,T\) 

image

Induced emf across each spoke is given by 

\(\mathcal{E} = \dfrac{1}{2} B_{earth} \,\omega\ell^2\)

   \(= \dfrac{1}{2} (0.5× 10^{-4}) (1) (10)^2\)

   \(= 0.25× 10^{-4} × 100\) 

   \(= 2.5× 10^{-3} \,V\)

   \(= 2.5 \,mV\)

image

A Ferris Wheel  has radial metallic spokes between the hub and the circular rim. These spokes move in the magnetic field of the earth. If the length of each spoke is \(\ell = \) 10 m and angular velocity is \(\omega = \) 1 rad/s, then find the induced emf across center of wheel and periphery. Given : Magnetic field of Earth, \(B _{earth} = 0.5 × 10^{-4 } \,T\)  (Assume that Bearth is horizontal at the location)

image
A

\(5 \,mV\)

.

B

\(2.5 \,mV\)

C

\(3 \,mV\)

D

\(4 \,mV\)

Option B is Correct

Illustration Questions

An aircraft is flying at \(v=\) 250 m/s over the region where earth's magnetic field points straight downwards. If the wing span of aircraft is \(L\)= 55 m, then find the induced motional emf between the wings. Given : Earth's magnetic field, \(B = 5.6 × 10^{-5} \,T\)  

A \(0.88\,V\)

B \(0.77\,V\)

C \(1.30\,V\)

D \(2.58\,V\)

×

Velocity of aircraft, \(v= 250\,m/s\)

Distance between wings, \(L=55 \,m\) 

Earth's magnetic field, \(B\) = \(5.6 × 10^{-5} \,T\)

image

The induced motional emf between the wings is given by 

\(\mathcal{E} = B \,v\,L\)

\(\mathcal{E} = 5.6 × 10^{-5} × 250 × 55\)

\(\mathcal{E} = 77000× 10^{-5} \)

\(\mathcal{E} = 0.77\, V\)

image

An aircraft is flying at \(v=\) 250 m/s over the region where earth's magnetic field points straight downwards. If the wing span of aircraft is \(L\)= 55 m, then find the induced motional emf between the wings. Given : Earth's magnetic field, \(B = 5.6 × 10^{-5} \,T\)  

image
A

\(0.88\,V\)

.

B

\(0.77\,V\)

C

\(1.30\,V\)

D

\(2.58\,V\)

Option B is Correct

Practice Now