6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

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Payal Gupta | Contributor-Level 10

4 months ago

6.4 Length of the rectangular wire loop, l = 8 cm = 0.08 m

Width of the rectangular loop, b = 2 cm = 0.02 m

Hence, the area of the rectangular loop, A = l $\times b$ = 0.08 $\times 0.02$ = 1.6 $\times 10^{- 3}$ $m^{2}$

Magnetic field strength, B= 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m /s

Emf developed in the longer side:

Emf developed in the loop is given by e = Blv = 0.3 $\times 0.08 \times 0.01$ = 2.4 $\times 10^{- 4}$ V

Time taken to travel the width = $\frac{D i s t a n c e t r a v e l l e d}{v e l o c i t y}$ = $\frac{b}{v}$ = $\frac{0.02}{0.01}$ s= 2 s

Hence the induced voltage is 2.4 $\times 10^{- 4}$ V, which lasts for 2 s.

Emf developed in the shorter side:

Emf developed in the loop is given by e = Bbv = 0.3 $\times 0.02 \times 0.01$ = 0.6 $\times 10^{- 4}$ &n

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6.4 Length of the rectangular wire loop, l = 8 cm = 0.08 m

Width of the rectangular loop, b = 2 cm = 0.02 m

Hence, the area of the rectangular loop, A = l $\times b$ = 0.08 $\times 0.02$ = 1.6 $\times 10^{- 3}$ $m^{2}$

Magnetic field strength, B= 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m /s

Emf developed in the longer side:

Emf developed in the loop is given by e = Blv = 0.3 $\times 0.08 \times 0.01$ = 2.4 $\times 10^{- 4}$ V

Time taken to travel the width = $\frac{D i s t a n c e t r a v e l l e d}{v e l o c i t y}$ = $\frac{b}{v}$ = $\frac{0.02}{0.01}$ s= 2 s

Hence the induced voltage is 2.4 $\times 10^{- 4}$ V, which lasts for 2 s.

Emf developed in the shorter side:

Emf developed in the loop is given by e = Bbv = 0.3 $\times 0.02 \times 0.01$ = 0.6 $\times 10^{- 4}$ V

Time taken to travel the width = $\frac{D i s t a n c e t r a v e l l e d}{v e l o c i t y}$ = $\frac{l}{v}$ = $\frac{0.08}{0.01}$ s= 8 s

Hence the induced voltage is 0.6 $\times 10^{- 4}$ V, which lasts for 8 s.

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6.4 Length of the rectangular wire loop, l = 8 cm = 0.08 mWidth of the rectangular loop, b = 2 cm = 0.02 mHence, the area of the rectangular loop, A = l <math><mo>×</mo><mi>b</mi></math> = 0.08 <math><mo>×</mo><mn>0.02</mn></math> = 1.6 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math>Magnetic field strength, B= 0.3 TVelocity of the loop, v = 1 cm/s = 0.01 m /sEmf developed in the longer side:Emf developed in the loop is given by e = Blv = 0.3 <math><mo>×</mo><mn>0.08</mn><mo>×</mo><mn>0.01</mn></math> = 2.4 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math> VTime taken to travel the width = <math><mfrac><mrow><mrow><mi>D</mi><mi>i</mi><mi>s</mi><mi>t</mi><mi>a</mi><mi>n</mi><mi>c</mi><mi>e</mi><mi></mi><mi>t</mi><mi>r</mi><mi>a</mi><mi>v</mi><mi>e</mi><mi>l</mi><mi>l</mi><mi>e</mi><mi>d</mi></mrow></mrow><mrow><mrow><mi>v</mi><mi>e</mi><mi>l</mi><mi>o</mi><mi>c</mi><mi>i</mi><mi>t</mi><mi>y</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mi>v</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>0.02</mn></mrow></mrow><mrow><mrow><mn>0.01</mn></mrow></mrow></mfrac></math> s= 2 sHence the induced voltage is 2.4 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math> V, which lasts for 2 s.Emf developed in the shorter side:Emf developed in the loop is given by e = Bbv = 0.3 <math><mo>×</mo><mn>0.02</mn><mo>×</mo><mn>0.01</mn></math> = 0.6 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math> VTime taken to travel the width = <math><mfrac><mrow><mrow><mi>D</mi><mi>i</mi><mi>s</mi><mi>t</mi><mi>a</mi><mi>n</mi><mi>c</mi><mi>e</mi><mi></mi><mi>t</mi><mi>r</mi><mi>a</mi><mi>v</mi><mi>e</mi><mi>l</mi><mi>l</mi><mi>e</mi><mi>d</mi></mrow></mrow><mrow><mrow><mi>v</mi><mi>e</mi><mi>l</mi><mi>o</mi><mi>c</mi><mi>i</mi><mi>t</mi><mi>y</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mi>v</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>0.08</mn></mrow></mrow><mrow><mrow><mn>0.01</mn></mrow></mrow></mfrac></math> s= 8 sHence the induced voltage is 0.6 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math> V, which lasts for 8 s.

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