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Moving Charges & Magnetism

153553 Assertion: Magnetic field cannot change K.E. of moving charge.
Reason: Magnetic field cannot change velocity vector.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect.

AIIMS-26.05.2018(M)

Moving Charges & Magnetism

153554 A beam of protons with velocity $600 km / s$ enters a uniform magnetic field of $0.2 T$ . the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be

1

1.56 cm

2

2.56 cm

3

3.56 cm

4

4.56 cm

Explanation:

A Given,
$\Rightarrow v = 600 km / \sec$
$v = 6 \times 10^{5} m / \sec$
$B = 0.2 T$
$θ = 30^{\circ}$
We know,
$m_{p} = 1.67 \times 10^{- 27} kg$
$q = 1.6 \times 10^{- 19} C$
The Radius of helical path,
$r = \frac{m v \sin θ}{qB} = \frac{1.67 \times 10^{- 27} \times 6 \times 10^{5} \sin 30^{\circ}}{1.6 \times 10^{- 19} \times 0.2}$
$r = \frac{1.67 \times 6 \times 10^{- 3}}{1.6 \times 0.2 \times 2}$
$r = 1.56 cm .$

Assam CEE-2018

Moving Charges & Magnetism

153555 An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$ . Suddenly, the field is reduced to $\frac{B}{2}$ . The radius of the circular path now becomes :

1

\frac{r}{2}

2

2 r

3

\frac{r}{4}

4

4 r

Explanation:

B Given, radius $= r$ , magnetic field $= B$
We know-
The radius of circular path $(r) = \frac{mv}{Bq}$
Here, $r \propto \frac{mv}{B}$
Now magnetic field is reduced to $\frac{B}{2}$ .
So, new magnetic field $(B^{'}) = \frac{B}{2}$ .
Now, Radius of new path $(r^{'}) = \frac{mv}{Bq}$
$r^{'} = \frac{m v}{\frac{B}{2} q}, r^{'} = \frac{2 m v}{B q}$
$r^{'} = 2 r$
Hence, new radius of circular path is $2 r$ .

Karnataka CET-2018

Moving Charges & Magnetism

153556 A charge $q$ is accelerated through a potential difference $V$ . It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$ . The potential difference required to move it in a circle of radius $2 r$ is :

1

2 V

2

4 V

3

1 V

4

3 V

Explanation:

B Given, $r_{1} = 2 r$
Radius of circular path $(r) = \frac{mv}{qB} = \frac{\sqrt{2 mqV}}{qB}$
$r \propto \sqrt{V} [∵ B = Constant]$
$or r^{2} \propto V$
$So, \frac{V_{1}}{V} = {(\frac{r_{1}}{r})}^{2} = {(\frac{2 r}{r})}^{2} = 4$
$or V_{1} = 4 V$

Karnataka CET-2018

Moving Charges & Magnetism

153553 Assertion: Magnetic field cannot change K.E. of moving charge.
Reason: Magnetic field cannot change velocity vector.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect.

AIIMS-26.05.2018(M)

Moving Charges & Magnetism

153554 A beam of protons with velocity $600 km / s$ enters a uniform magnetic field of $0.2 T$ . the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be

1

1.56 cm

2

2.56 cm

3

3.56 cm

4

4.56 cm

Explanation:

A Given,
$\Rightarrow v = 600 km / \sec$
$v = 6 \times 10^{5} m / \sec$
$B = 0.2 T$
$θ = 30^{\circ}$
We know,
$m_{p} = 1.67 \times 10^{- 27} kg$
$q = 1.6 \times 10^{- 19} C$
The Radius of helical path,
$r = \frac{m v \sin θ}{qB} = \frac{1.67 \times 10^{- 27} \times 6 \times 10^{5} \sin 30^{\circ}}{1.6 \times 10^{- 19} \times 0.2}$
$r = \frac{1.67 \times 6 \times 10^{- 3}}{1.6 \times 0.2 \times 2}$
$r = 1.56 cm .$

Assam CEE-2018

Moving Charges & Magnetism

153555 An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$ . Suddenly, the field is reduced to $\frac{B}{2}$ . The radius of the circular path now becomes :

1

\frac{r}{2}

2

2 r

3

\frac{r}{4}

4

4 r

Explanation:

B Given, radius $= r$ , magnetic field $= B$
We know-
The radius of circular path $(r) = \frac{mv}{Bq}$
Here, $r \propto \frac{mv}{B}$
Now magnetic field is reduced to $\frac{B}{2}$ .
So, new magnetic field $(B^{'}) = \frac{B}{2}$ .
Now, Radius of new path $(r^{'}) = \frac{mv}{Bq}$
$r^{'} = \frac{m v}{\frac{B}{2} q}, r^{'} = \frac{2 m v}{B q}$
$r^{'} = 2 r$
Hence, new radius of circular path is $2 r$ .

Karnataka CET-2018

Moving Charges & Magnetism

153556 A charge $q$ is accelerated through a potential difference $V$ . It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$ . The potential difference required to move it in a circle of radius $2 r$ is :

1

2 V

2

4 V

3

1 V

4

3 V

Explanation:

B Given, $r_{1} = 2 r$
Radius of circular path $(r) = \frac{mv}{qB} = \frac{\sqrt{2 mqV}}{qB}$
$r \propto \sqrt{V} [∵ B = Constant]$
$or r^{2} \propto V$
$So, \frac{V_{1}}{V} = {(\frac{r_{1}}{r})}^{2} = {(\frac{2 r}{r})}^{2} = 4$
$or V_{1} = 4 V$

Karnataka CET-2018

Moving Charges & Magnetism

153553 Assertion: Magnetic field cannot change K.E. of moving charge.
Reason: Magnetic field cannot change velocity vector.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect.

AIIMS-26.05.2018(M)

Moving Charges & Magnetism

153554 A beam of protons with velocity $600 km / s$ enters a uniform magnetic field of $0.2 T$ . the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be

1

1.56 cm

2

2.56 cm

3

3.56 cm

4

4.56 cm

Explanation:

A Given,
$\Rightarrow v = 600 km / \sec$
$v = 6 \times 10^{5} m / \sec$
$B = 0.2 T$
$θ = 30^{\circ}$
We know,
$m_{p} = 1.67 \times 10^{- 27} kg$
$q = 1.6 \times 10^{- 19} C$
The Radius of helical path,
$r = \frac{m v \sin θ}{qB} = \frac{1.67 \times 10^{- 27} \times 6 \times 10^{5} \sin 30^{\circ}}{1.6 \times 10^{- 19} \times 0.2}$
$r = \frac{1.67 \times 6 \times 10^{- 3}}{1.6 \times 0.2 \times 2}$
$r = 1.56 cm .$

Assam CEE-2018

Moving Charges & Magnetism

153555 An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$ . Suddenly, the field is reduced to $\frac{B}{2}$ . The radius of the circular path now becomes :

1

\frac{r}{2}

2

2 r

3

\frac{r}{4}

4

4 r

Explanation:

B Given, radius $= r$ , magnetic field $= B$
We know-
The radius of circular path $(r) = \frac{mv}{Bq}$
Here, $r \propto \frac{mv}{B}$
Now magnetic field is reduced to $\frac{B}{2}$ .
So, new magnetic field $(B^{'}) = \frac{B}{2}$ .
Now, Radius of new path $(r^{'}) = \frac{mv}{Bq}$
$r^{'} = \frac{m v}{\frac{B}{2} q}, r^{'} = \frac{2 m v}{B q}$
$r^{'} = 2 r$
Hence, new radius of circular path is $2 r$ .

Karnataka CET-2018

Moving Charges & Magnetism

153556 A charge $q$ is accelerated through a potential difference $V$ . It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$ . The potential difference required to move it in a circle of radius $2 r$ is :

1

2 V

2

4 V

3

1 V

4

3 V

Explanation:

B Given, $r_{1} = 2 r$
Radius of circular path $(r) = \frac{mv}{qB} = \frac{\sqrt{2 mqV}}{qB}$
$r \propto \sqrt{V} [∵ B = Constant]$
$or r^{2} \propto V$
$So, \frac{V_{1}}{V} = {(\frac{r_{1}}{r})}^{2} = {(\frac{2 r}{r})}^{2} = 4$
$or V_{1} = 4 V$

Karnataka CET-2018

Moving Charges & Magnetism

153553 Assertion: Magnetic field cannot change K.E. of moving charge.
Reason: Magnetic field cannot change velocity vector.

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect.

AIIMS-26.05.2018(M)

Moving Charges & Magnetism

153554 A beam of protons with velocity $600 km / s$ enters a uniform magnetic field of $0.2 T$ . the velocity make an angle of $30^{\circ}$ with the magnetic field. The radius of the helical path will be

1

1.56 cm

2

2.56 cm

3

3.56 cm

4

4.56 cm

Explanation:

A Given,
$\Rightarrow v = 600 km / \sec$
$v = 6 \times 10^{5} m / \sec$
$B = 0.2 T$
$θ = 30^{\circ}$
We know,
$m_{p} = 1.67 \times 10^{- 27} kg$
$q = 1.6 \times 10^{- 19} C$
The Radius of helical path,
$r = \frac{m v \sin θ}{qB} = \frac{1.67 \times 10^{- 27} \times 6 \times 10^{5} \sin 30^{\circ}}{1.6 \times 10^{- 19} \times 0.2}$
$r = \frac{1.67 \times 6 \times 10^{- 3}}{1.6 \times 0.2 \times 2}$
$r = 1.56 cm .$

Assam CEE-2018

Moving Charges & Magnetism

153555 An electron is moving in a circle of radius $r$ in a uniform magnetic field $B$ . Suddenly, the field is reduced to $\frac{B}{2}$ . The radius of the circular path now becomes :

1

\frac{r}{2}

2

2 r

3

\frac{r}{4}

4

4 r

Explanation:

B Given, radius $= r$ , magnetic field $= B$
We know-
The radius of circular path $(r) = \frac{mv}{Bq}$
Here, $r \propto \frac{mv}{B}$
Now magnetic field is reduced to $\frac{B}{2}$ .
So, new magnetic field $(B^{'}) = \frac{B}{2}$ .
Now, Radius of new path $(r^{'}) = \frac{mv}{Bq}$
$r^{'} = \frac{m v}{\frac{B}{2} q}, r^{'} = \frac{2 m v}{B q}$
$r^{'} = 2 r$
Hence, new radius of circular path is $2 r$ .

Karnataka CET-2018

Moving Charges & Magnetism

153556 A charge $q$ is accelerated through a potential difference $V$ . It is then passed normally through a uniform magnetic field, where it moves in a circle of radius $r$ . The potential difference required to move it in a circle of radius $2 r$ is :

1

2 V

2

4 V

3

1 V

4

3 V

Explanation:

B Given, $r_{1} = 2 r$
Radius of circular path $(r) = \frac{mv}{qB} = \frac{\sqrt{2 mqV}}{qB}$
$r \propto \sqrt{V} [∵ B = Constant]$
$or r^{2} \propto V$
$So, \frac{V_{1}}{V} = {(\frac{r_{1}}{r})}^{2} = {(\frac{2 r}{r})}^{2} = 4$
$or V_{1} = 4 V$

Karnataka CET-2018