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This is the second of our many quizzes on Faraday's law. Click on this to access more such competitive quizzes on Current Electricity and EMI.

Faraday's law Quiz 2

physics

 A series of LEDs is connected to a coil of wire which is placed in a magnetic field. In which of the following cases the LEDs will not glow? 

The Induced emf is given by, $\in =-\dfrac{\Delta {{\phi }_{B}}}{\Delta t}$. 

  To induce a current or an emf, it requires a change in the magnetic flux. Only then the current is induced. Induced emf is, $\in =-\dfrac{\Delta {{\phi }_{B}}}{\Delta t}$. So, to not induce any current or to not make the bulb glow, the magnetic flux should not have to change. When the bulb is kept in the magnetic field without moving, the magnetic flux will not change, hence, the current is not induced. So, the bulb will not glow.

 A rectangular coil of length 4 cm and breadth 3.5 cm is been placed parallel to a magnetic field of strength 2.5T. Calculate the magnetic flux of the coil.$\left( \theta =0 \right)$<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/dccf9974-9278-4e52-811a-c68eaa862ab1833916967199080197.png"/>

The magnetic flux is calculated by the formula, ${{\phi }_{B}}=BA\cos \theta $. Taking the area of the rectangular loop as the product of length and breadth. 

  $\text{W.K.T the magnetic flux is given by,}\\{{\phi }_{B}}=BA\cos \theta\\\text{Since, it is a rectangular coil,}\\A=l\left( b \right)\\\text{Now, magnetic flux is calculated as,}\\\Rightarrow{{\phi }_{B}}=B\left( l \right)\left( b \right)\cos 0\\\Rightarrow {{\phi }_{B}}=2.5\left( 0.04 \right)\left( 0.035 \right) \\\therefore{{\phi }_{B}}=0.0035\,Wb$

 A rectangular coil of breadth 5 cm has been placed inside the magnetic field of strength $0.6 T$.It has been removed over a time of 0.6 seconds. The induced emf produced is equal to $0.034V$. Calculate the length of the rectangular loop in cm, taking $\left( \theta =0 \right)$  

The induced emf is found by the formula, $\varepsilon =-\dfrac{\Delta \phi }{\Delta t}$. The magnetic flux is given by the formula, ${{\phi }_{B}}=BA\cos \theta $.  

 $\text{W.K.T, the induced emf is given by,}\\\varepsilon =-\dfrac{\Delta \phi }{\Delta t}\\\Rightarrow 0.034=-\dfrac{\Delta \phi }{0.6} \\\Rightarrow -\Delta \phi =0.0204 \\\Rightarrow {{\phi }_{f}}-{{\phi }_{i}}=\left( 0-0.0204 \right) \\\Rightarrow 0-{{\phi }_{i}}=0-0.024 \\\Rightarrow {{\phi }_{i}}=0.0204 \\\text{W.K.T the magnetic flux is given by the formula,}\\{{\phi }_{B}}=B\left( l \right)\left( b \right)\cos 0\,..........\text{(For a rectangular loop the area=lb)}\\\Rightarrow 0.0204=0.6\left( l \right)0.05\\\Rightarrow l=\dfrac{0.0204}{0.03} \\\Rightarrow l=0.68m \\\therefore l=68\,cm$ 

A circular loop of wire is placed perpendicular to the magnetic field of strength $0.5T$. Calculate the magnetic flux of the wire in Wb if the radius of the loop is $3cm$. 

The magnetic flux is calculated by the formula, ${{\phi }_{B}}=BA\cos \theta $.

 We know that, magnetic flux is calculated as $\,{{\phi }_{B}}=BA\cos \theta\\ $ Since, the loop is perpendicular to the magnetic field, then $\theta =90{}^\circ$ $\\{{\phi }_{B}}=0.5\left( \pi {{\left( 0.03 \right)}^{2}} \right) \times 0 $ This implies that the magnetic flux inside the loop is cancelled and it is equal to zero.

 Calculate the induced emf, when the rectangular loop of length $9\,cm$ and breadth $5\,cm$ is removed from a magnetic field of strength 3.5T to a position where magnetic flux is equal to $0.45\,Wb$ over a time of $0.45$ seconds.$(\text{ Take}\,\theta =0)$ 

The magnetic flux is obtained by the formula, ${{\phi }_{B}}=BA\cos \theta$. The induced emf is found by the formula, $\varepsilon =-\dfrac{\Delta \phi }{\Delta t}$. 

 $\text{The initial magnetic flux is calculated by the formula,}\\{{\phi }_{{{B}_{i}}}}=BA\cos \theta\\\Rightarrow{{\phi }_{{{B}_{i}}}}=3.5\left( 0.09 \right)0.05 \\\Rightarrow {{\phi }_{{{B}_{i}}}}=0.01575Wb \\\Rightarrow{{\phi }_{{{B}_{f}}}}=0.45Wb\\\text{Now the change in the magnetic flux is,}\\\Delta {{\phi }_{B}}=\left( 0.45-0.01575 \right)\\\Rightarrow\Delta {{\phi }_{B}}=0.43425\,Wb\\\text{Now, the induced emf is found by the formula,}\\\varepsilon =-\dfrac{\Delta \phi }{\Delta t}\\\Rightarrow\varepsilon =\dfrac{-0.43425}{0.45} \\\therefore\varepsilon =-0.965\,V$