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Viktor [21]
3 years ago
10

A conducting rod moves perpendicularly through a uniform magnetic field. How does the emf change if the magnetic field is increa

sed by a factor of 4 and the time required for the rod to move through the field is decreased to half of the original time?
a) It increases by a factor of 2
b) It increases by a factor of 4
c) It increases by a factor of 8
d) It increases by a factor of 16 .
Physics
1 answer:
Marysya12 [62]3 years ago
5 0

Answer:

c) It increases by a factor of 8

Explanation:

According to Faraday's law (and Lenz' law), the induced EMF is given as the rate of change of magnetic flux.

Mathematically:

V = -dФ/dt

Magnetic flux, Ф, is given as:

Ф = BA

where B  = magnetic field strength and A = Area of object

Hence, induced EMF becomes:

V = -d(BA)/dt  or -BA/t

If the magnetic field is increased by a factor of 4, (B_n = 4B) and the time required for the rod to move is decreased by a factor of 2 (t_n = t/2), the induced EMF becomes:

V_n = -(B_nA)/t_n

V_n = \frac{-4BA}{(t/2)}\\\\V_n = \frac{-8BA}{t} \\\\V_n = 8V\\

The EMF has increased by a factor of 8.

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How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un
Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

W=K.E

W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2

Put the value into the formula

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=1.122\times10^{-9}\ J

W=7001\ MeV

(b). We need to calculate  the momentum of this proton

Using formula of momentum

p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value into the formula

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}

p=1.404\times10^{-26}c

p=4.20\times10^{8}\ kg-m/s

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

4 0
3 years ago
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