153563
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$. If the current and radius each of them are made three times, the new ratio will become
1 $3 x$
2 $9 \mathrm{x}$
3 $\mathrm{x} / 9$
4 $\mathrm{x} / 27$
Explanation:
D Given, ratio $x=\frac{B}{M}=\frac{\mu_{0}}{2 \pi r^{3}}$ Magnetic field at a center of current carrying coil $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ The magnetic moment at the centre of a current carrying circular coil is given by, $\mathrm{M}=\mathrm{IA}=\mathrm{I} \pi \mathrm{r}^{2}$ If current and radius each of them are made three times, $x^{\prime}=\frac{\mu_{0}}{2 \pi(3 r)^{3}}$ $x^{\prime}=\frac{\mu_{o}}{2 \pi r^{3}} \times \frac{1}{27}$ So, $\quad x^{\prime}=\frac{x}{27}$
UPSEE - 2018
Moving Charges & Magnetism
153564
A non - conducting ring carries linear charge density $\lambda$. It is rotating with angular speed $\omega$ about its axis. The magnetic field at its centre is
1 $\frac{3 \mu_{0} \lambda \omega}{2 \pi}$
2 $\frac{\mu_{0} \lambda \omega}{2}$
3 $\frac{\mu_{0} \lambda \omega}{\pi}$
4 $\mu_{0} \lambda \omega$
Explanation:
B Given, linear charge density $=\lambda$, angular speed $=\omega$ We know that, Liner charge density $(\lambda)=\frac{\mathrm{q}}{2 \pi \mathrm{r}}$ $\mathrm{q}=2 \pi \mathrm{r} \lambda$ The magnetic field at the centre of ring $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}$ Now, the current is charge per unit time $\mathrm{i}=\mathrm{q} / \mathrm{t}=\frac{\mathrm{q}}{2 \pi / \omega}=\frac{\mathrm{q} \omega}{2 \pi} \quad\left(\because \mathrm{t}=\frac{2 \pi}{\omega}\right)$ $\mathrm{i}=\frac{2 \pi \mathrm{r} \lambda \omega}{2 \pi}=\mathrm{r} \lambda \omega$ Substituting equation (ii) in equation (i), we get - $\mathrm{B}=\frac{\mu_{0} \lambda \omega}{2}$
UPSEE - 2018
Moving Charges & Magnetism
153565
A charged particle travels along a straight line with a speed $v$ in a region where both electric field $E$ and magnetic fields $B$ are present. It follows that
1 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are parallel
2 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are perpendicular
3 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are parallel
4 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are perpendicular
Explanation:
B Given, speed $=\mathrm{v}$, electric field $=\mathrm{E}$, magnetic field $=\mathrm{B}$ We know, Force due to magnetic field is - $\mathrm{F}=\mathrm{qvB} \sin \theta \text {. }$ When, $\theta=90^{\circ}$ $\mathrm{F}=\mathrm{qvB}$ Force due to electric field $\mathrm{E}$ is $\mathrm{F}=\mathrm{qE}$ From equation (i) and equation (ii), we get - $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ Hence, $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two field are perpendicular.
Manipal UGET-2018
Moving Charges & Magnetism
153566
An $\alpha$-particle moving with velocity $5 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$ enters in a magnetic field $(3 \hat{i}+2 \hat{j}) T$. The force experienced by the particle is
153563
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$. If the current and radius each of them are made three times, the new ratio will become
1 $3 x$
2 $9 \mathrm{x}$
3 $\mathrm{x} / 9$
4 $\mathrm{x} / 27$
Explanation:
D Given, ratio $x=\frac{B}{M}=\frac{\mu_{0}}{2 \pi r^{3}}$ Magnetic field at a center of current carrying coil $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ The magnetic moment at the centre of a current carrying circular coil is given by, $\mathrm{M}=\mathrm{IA}=\mathrm{I} \pi \mathrm{r}^{2}$ If current and radius each of them are made three times, $x^{\prime}=\frac{\mu_{0}}{2 \pi(3 r)^{3}}$ $x^{\prime}=\frac{\mu_{o}}{2 \pi r^{3}} \times \frac{1}{27}$ So, $\quad x^{\prime}=\frac{x}{27}$
UPSEE - 2018
Moving Charges & Magnetism
153564
A non - conducting ring carries linear charge density $\lambda$. It is rotating with angular speed $\omega$ about its axis. The magnetic field at its centre is
1 $\frac{3 \mu_{0} \lambda \omega}{2 \pi}$
2 $\frac{\mu_{0} \lambda \omega}{2}$
3 $\frac{\mu_{0} \lambda \omega}{\pi}$
4 $\mu_{0} \lambda \omega$
Explanation:
B Given, linear charge density $=\lambda$, angular speed $=\omega$ We know that, Liner charge density $(\lambda)=\frac{\mathrm{q}}{2 \pi \mathrm{r}}$ $\mathrm{q}=2 \pi \mathrm{r} \lambda$ The magnetic field at the centre of ring $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}$ Now, the current is charge per unit time $\mathrm{i}=\mathrm{q} / \mathrm{t}=\frac{\mathrm{q}}{2 \pi / \omega}=\frac{\mathrm{q} \omega}{2 \pi} \quad\left(\because \mathrm{t}=\frac{2 \pi}{\omega}\right)$ $\mathrm{i}=\frac{2 \pi \mathrm{r} \lambda \omega}{2 \pi}=\mathrm{r} \lambda \omega$ Substituting equation (ii) in equation (i), we get - $\mathrm{B}=\frac{\mu_{0} \lambda \omega}{2}$
UPSEE - 2018
Moving Charges & Magnetism
153565
A charged particle travels along a straight line with a speed $v$ in a region where both electric field $E$ and magnetic fields $B$ are present. It follows that
1 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are parallel
2 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are perpendicular
3 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are parallel
4 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are perpendicular
Explanation:
B Given, speed $=\mathrm{v}$, electric field $=\mathrm{E}$, magnetic field $=\mathrm{B}$ We know, Force due to magnetic field is - $\mathrm{F}=\mathrm{qvB} \sin \theta \text {. }$ When, $\theta=90^{\circ}$ $\mathrm{F}=\mathrm{qvB}$ Force due to electric field $\mathrm{E}$ is $\mathrm{F}=\mathrm{qE}$ From equation (i) and equation (ii), we get - $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ Hence, $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two field are perpendicular.
Manipal UGET-2018
Moving Charges & Magnetism
153566
An $\alpha$-particle moving with velocity $5 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$ enters in a magnetic field $(3 \hat{i}+2 \hat{j}) T$. The force experienced by the particle is
153563
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$. If the current and radius each of them are made three times, the new ratio will become
1 $3 x$
2 $9 \mathrm{x}$
3 $\mathrm{x} / 9$
4 $\mathrm{x} / 27$
Explanation:
D Given, ratio $x=\frac{B}{M}=\frac{\mu_{0}}{2 \pi r^{3}}$ Magnetic field at a center of current carrying coil $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ The magnetic moment at the centre of a current carrying circular coil is given by, $\mathrm{M}=\mathrm{IA}=\mathrm{I} \pi \mathrm{r}^{2}$ If current and radius each of them are made three times, $x^{\prime}=\frac{\mu_{0}}{2 \pi(3 r)^{3}}$ $x^{\prime}=\frac{\mu_{o}}{2 \pi r^{3}} \times \frac{1}{27}$ So, $\quad x^{\prime}=\frac{x}{27}$
UPSEE - 2018
Moving Charges & Magnetism
153564
A non - conducting ring carries linear charge density $\lambda$. It is rotating with angular speed $\omega$ about its axis. The magnetic field at its centre is
1 $\frac{3 \mu_{0} \lambda \omega}{2 \pi}$
2 $\frac{\mu_{0} \lambda \omega}{2}$
3 $\frac{\mu_{0} \lambda \omega}{\pi}$
4 $\mu_{0} \lambda \omega$
Explanation:
B Given, linear charge density $=\lambda$, angular speed $=\omega$ We know that, Liner charge density $(\lambda)=\frac{\mathrm{q}}{2 \pi \mathrm{r}}$ $\mathrm{q}=2 \pi \mathrm{r} \lambda$ The magnetic field at the centre of ring $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}$ Now, the current is charge per unit time $\mathrm{i}=\mathrm{q} / \mathrm{t}=\frac{\mathrm{q}}{2 \pi / \omega}=\frac{\mathrm{q} \omega}{2 \pi} \quad\left(\because \mathrm{t}=\frac{2 \pi}{\omega}\right)$ $\mathrm{i}=\frac{2 \pi \mathrm{r} \lambda \omega}{2 \pi}=\mathrm{r} \lambda \omega$ Substituting equation (ii) in equation (i), we get - $\mathrm{B}=\frac{\mu_{0} \lambda \omega}{2}$
UPSEE - 2018
Moving Charges & Magnetism
153565
A charged particle travels along a straight line with a speed $v$ in a region where both electric field $E$ and magnetic fields $B$ are present. It follows that
1 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are parallel
2 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are perpendicular
3 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are parallel
4 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are perpendicular
Explanation:
B Given, speed $=\mathrm{v}$, electric field $=\mathrm{E}$, magnetic field $=\mathrm{B}$ We know, Force due to magnetic field is - $\mathrm{F}=\mathrm{qvB} \sin \theta \text {. }$ When, $\theta=90^{\circ}$ $\mathrm{F}=\mathrm{qvB}$ Force due to electric field $\mathrm{E}$ is $\mathrm{F}=\mathrm{qE}$ From equation (i) and equation (ii), we get - $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ Hence, $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two field are perpendicular.
Manipal UGET-2018
Moving Charges & Magnetism
153566
An $\alpha$-particle moving with velocity $5 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$ enters in a magnetic field $(3 \hat{i}+2 \hat{j}) T$. The force experienced by the particle is
153563
The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is $x$. If the current and radius each of them are made three times, the new ratio will become
1 $3 x$
2 $9 \mathrm{x}$
3 $\mathrm{x} / 9$
4 $\mathrm{x} / 27$
Explanation:
D Given, ratio $x=\frac{B}{M}=\frac{\mu_{0}}{2 \pi r^{3}}$ Magnetic field at a center of current carrying coil $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ The magnetic moment at the centre of a current carrying circular coil is given by, $\mathrm{M}=\mathrm{IA}=\mathrm{I} \pi \mathrm{r}^{2}$ If current and radius each of them are made three times, $x^{\prime}=\frac{\mu_{0}}{2 \pi(3 r)^{3}}$ $x^{\prime}=\frac{\mu_{o}}{2 \pi r^{3}} \times \frac{1}{27}$ So, $\quad x^{\prime}=\frac{x}{27}$
UPSEE - 2018
Moving Charges & Magnetism
153564
A non - conducting ring carries linear charge density $\lambda$. It is rotating with angular speed $\omega$ about its axis. The magnetic field at its centre is
1 $\frac{3 \mu_{0} \lambda \omega}{2 \pi}$
2 $\frac{\mu_{0} \lambda \omega}{2}$
3 $\frac{\mu_{0} \lambda \omega}{\pi}$
4 $\mu_{0} \lambda \omega$
Explanation:
B Given, linear charge density $=\lambda$, angular speed $=\omega$ We know that, Liner charge density $(\lambda)=\frac{\mathrm{q}}{2 \pi \mathrm{r}}$ $\mathrm{q}=2 \pi \mathrm{r} \lambda$ The magnetic field at the centre of ring $\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}$ Now, the current is charge per unit time $\mathrm{i}=\mathrm{q} / \mathrm{t}=\frac{\mathrm{q}}{2 \pi / \omega}=\frac{\mathrm{q} \omega}{2 \pi} \quad\left(\because \mathrm{t}=\frac{2 \pi}{\omega}\right)$ $\mathrm{i}=\frac{2 \pi \mathrm{r} \lambda \omega}{2 \pi}=\mathrm{r} \lambda \omega$ Substituting equation (ii) in equation (i), we get - $\mathrm{B}=\frac{\mu_{0} \lambda \omega}{2}$
UPSEE - 2018
Moving Charges & Magnetism
153565
A charged particle travels along a straight line with a speed $v$ in a region where both electric field $E$ and magnetic fields $B$ are present. It follows that
1 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are parallel
2 $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two fields are perpendicular
3 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are parallel
4 $|\mathrm{B}|=\mathrm{v}|\mathrm{E}|$ and the two fields are perpendicular
Explanation:
B Given, speed $=\mathrm{v}$, electric field $=\mathrm{E}$, magnetic field $=\mathrm{B}$ We know, Force due to magnetic field is - $\mathrm{F}=\mathrm{qvB} \sin \theta \text {. }$ When, $\theta=90^{\circ}$ $\mathrm{F}=\mathrm{qvB}$ Force due to electric field $\mathrm{E}$ is $\mathrm{F}=\mathrm{qE}$ From equation (i) and equation (ii), we get - $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ Hence, $|\mathrm{E}|=\mathrm{v}|\mathrm{B}|$ and the two field are perpendicular.
Manipal UGET-2018
Moving Charges & Magnetism
153566
An $\alpha$-particle moving with velocity $5 \times 10^{5} \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}$ enters in a magnetic field $(3 \hat{i}+2 \hat{j}) T$. The force experienced by the particle is
