153314
A one metre long wire is lying at right angles to the magnetic field. A force of $1 \mathrm{~kg} \mathrm{wt}$. is acting on it in a magnetic field of 0.98 tesla. The current flowing in it will be
1 $100 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 zero
Explanation:
B Given that $\mathrm{B}=0.98 \mathrm{~T}$ $l=1 \mathrm{~m}$ $\mathrm{~F}=1 \mathrm{~kg} \mathrm{wt}=1 \times 9.8 \mathrm{~N}$ $\theta=90^{\circ}$ We know that, Magnetic force $(F)=i B l \operatorname{Sin} \theta$ $\mathrm{F}=\mathrm{i} \mathrm{B} l \sin 90^{\circ}$ $\mathrm{F}=\mathrm{iB} l$ $1 \times 9.8=\mathrm{i} \times 0.98 \times 1$ $\mathrm{i}=10 \mathrm{amp} .$
J and K CET- 2004
Moving Charges & Magnetism
153316
A straight thin conductor is bent as shown in the adjoining figure. It carries a current $I$ ampere. The radius of the semicircular arc is $r$ meter. The magnetic induction at the centre of semicircular arc is
1 $\frac{\mu_{0}}{4 \pi} \frac{I}{r}$ tesla
2 $\frac{\mu_{0}}{4} \frac{I}{r}$ tesla
3 $\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{2 \mathrm{r}}$ tesla
4 zero
Explanation:
B As we know that, Magnetic field for a circular wire $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ For a semi circle Magnetic field $\left(\mathrm{B}^{\prime}\right)=\mathrm{B} / 2$ $\mathrm{B}^{\prime}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}} \text { Tesla }$
J and K CET- 2001
Moving Charges & Magnetism
153317
The magnetic field at a distance $r$ from a long wire carrying current $i$ is 0.6 Tesla. The magnetic field at distance $3 r$ is
1 $0.1 \mathrm{~T}$
2 $0.2 \mathrm{~T}$
3 $0.9 \mathrm{~T}$
4 $1.8 \mathrm{~T}$
Explanation:
B Given that $\mathrm{i}_{1}=\mathrm{i}_{2}=\mathrm{i} \quad \mathrm{B}_{1}=0.6 \mathrm{~T}$ $r_{1}=r, \quad r_{2}=3 r$ We know that magnetic field at a distance $r$ for a long wire carrying current $B=\frac{\mu_{0} i}{2 \pi r}$ $B \propto \frac{1}{r}$ Now ratio of magnetic field at distance $r_{1}$ and $r_{2}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}} =\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ $\frac{0.6}{\mathrm{~B}_{2}} =\left(\frac{3 \mathrm{r}}{\mathrm{r}}\right)$ $\mathrm{B}_{2} =\frac{0.6}{3}$ $\mathrm{~B}_{2} =0.2 \mathrm{~T}$
J and K CET- 1998
Moving Charges & Magnetism
153318
A long straight wire carries a current $10 \mathrm{~A}$ and has a length of $0.2 \mathrm{~m}$. The wire is positioned perpendicular to a magnetic field. If the wire experiences a force of $0.1 \mathrm{~N}$, the strength of the magnetic field is
1 $1.5 \mathrm{~T}$
2 $0.05 \mathrm{~T}$
3 $0.5 \mathrm{~T}$
4 zero
Explanation:
B Given that $\mathrm{i}=10 \mathrm{~A} \quad l=0.2 \mathrm{~m} \quad \theta=90^{\circ}$ $\mathrm{F}=0.1 \mathrm{~N}$ We know that $\mathrm{F}=\mathrm{iB} l \sin \theta$ $0.1=10 \times \mathrm{B} \times 0.2 \sin 90^{\circ}$ $0.1=10 \times \mathrm{B} \times 0.2 \times 1$ $\mathrm{~B}=0.05 \mathrm{~T}$
153314
A one metre long wire is lying at right angles to the magnetic field. A force of $1 \mathrm{~kg} \mathrm{wt}$. is acting on it in a magnetic field of 0.98 tesla. The current flowing in it will be
1 $100 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 zero
Explanation:
B Given that $\mathrm{B}=0.98 \mathrm{~T}$ $l=1 \mathrm{~m}$ $\mathrm{~F}=1 \mathrm{~kg} \mathrm{wt}=1 \times 9.8 \mathrm{~N}$ $\theta=90^{\circ}$ We know that, Magnetic force $(F)=i B l \operatorname{Sin} \theta$ $\mathrm{F}=\mathrm{i} \mathrm{B} l \sin 90^{\circ}$ $\mathrm{F}=\mathrm{iB} l$ $1 \times 9.8=\mathrm{i} \times 0.98 \times 1$ $\mathrm{i}=10 \mathrm{amp} .$
J and K CET- 2004
Moving Charges & Magnetism
153316
A straight thin conductor is bent as shown in the adjoining figure. It carries a current $I$ ampere. The radius of the semicircular arc is $r$ meter. The magnetic induction at the centre of semicircular arc is
1 $\frac{\mu_{0}}{4 \pi} \frac{I}{r}$ tesla
2 $\frac{\mu_{0}}{4} \frac{I}{r}$ tesla
3 $\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{2 \mathrm{r}}$ tesla
4 zero
Explanation:
B As we know that, Magnetic field for a circular wire $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ For a semi circle Magnetic field $\left(\mathrm{B}^{\prime}\right)=\mathrm{B} / 2$ $\mathrm{B}^{\prime}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}} \text { Tesla }$
J and K CET- 2001
Moving Charges & Magnetism
153317
The magnetic field at a distance $r$ from a long wire carrying current $i$ is 0.6 Tesla. The magnetic field at distance $3 r$ is
1 $0.1 \mathrm{~T}$
2 $0.2 \mathrm{~T}$
3 $0.9 \mathrm{~T}$
4 $1.8 \mathrm{~T}$
Explanation:
B Given that $\mathrm{i}_{1}=\mathrm{i}_{2}=\mathrm{i} \quad \mathrm{B}_{1}=0.6 \mathrm{~T}$ $r_{1}=r, \quad r_{2}=3 r$ We know that magnetic field at a distance $r$ for a long wire carrying current $B=\frac{\mu_{0} i}{2 \pi r}$ $B \propto \frac{1}{r}$ Now ratio of magnetic field at distance $r_{1}$ and $r_{2}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}} =\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ $\frac{0.6}{\mathrm{~B}_{2}} =\left(\frac{3 \mathrm{r}}{\mathrm{r}}\right)$ $\mathrm{B}_{2} =\frac{0.6}{3}$ $\mathrm{~B}_{2} =0.2 \mathrm{~T}$
J and K CET- 1998
Moving Charges & Magnetism
153318
A long straight wire carries a current $10 \mathrm{~A}$ and has a length of $0.2 \mathrm{~m}$. The wire is positioned perpendicular to a magnetic field. If the wire experiences a force of $0.1 \mathrm{~N}$, the strength of the magnetic field is
1 $1.5 \mathrm{~T}$
2 $0.05 \mathrm{~T}$
3 $0.5 \mathrm{~T}$
4 zero
Explanation:
B Given that $\mathrm{i}=10 \mathrm{~A} \quad l=0.2 \mathrm{~m} \quad \theta=90^{\circ}$ $\mathrm{F}=0.1 \mathrm{~N}$ We know that $\mathrm{F}=\mathrm{iB} l \sin \theta$ $0.1=10 \times \mathrm{B} \times 0.2 \sin 90^{\circ}$ $0.1=10 \times \mathrm{B} \times 0.2 \times 1$ $\mathrm{~B}=0.05 \mathrm{~T}$
153314
A one metre long wire is lying at right angles to the magnetic field. A force of $1 \mathrm{~kg} \mathrm{wt}$. is acting on it in a magnetic field of 0.98 tesla. The current flowing in it will be
1 $100 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 zero
Explanation:
B Given that $\mathrm{B}=0.98 \mathrm{~T}$ $l=1 \mathrm{~m}$ $\mathrm{~F}=1 \mathrm{~kg} \mathrm{wt}=1 \times 9.8 \mathrm{~N}$ $\theta=90^{\circ}$ We know that, Magnetic force $(F)=i B l \operatorname{Sin} \theta$ $\mathrm{F}=\mathrm{i} \mathrm{B} l \sin 90^{\circ}$ $\mathrm{F}=\mathrm{iB} l$ $1 \times 9.8=\mathrm{i} \times 0.98 \times 1$ $\mathrm{i}=10 \mathrm{amp} .$
J and K CET- 2004
Moving Charges & Magnetism
153316
A straight thin conductor is bent as shown in the adjoining figure. It carries a current $I$ ampere. The radius of the semicircular arc is $r$ meter. The magnetic induction at the centre of semicircular arc is
1 $\frac{\mu_{0}}{4 \pi} \frac{I}{r}$ tesla
2 $\frac{\mu_{0}}{4} \frac{I}{r}$ tesla
3 $\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{2 \mathrm{r}}$ tesla
4 zero
Explanation:
B As we know that, Magnetic field for a circular wire $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ For a semi circle Magnetic field $\left(\mathrm{B}^{\prime}\right)=\mathrm{B} / 2$ $\mathrm{B}^{\prime}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}} \text { Tesla }$
J and K CET- 2001
Moving Charges & Magnetism
153317
The magnetic field at a distance $r$ from a long wire carrying current $i$ is 0.6 Tesla. The magnetic field at distance $3 r$ is
1 $0.1 \mathrm{~T}$
2 $0.2 \mathrm{~T}$
3 $0.9 \mathrm{~T}$
4 $1.8 \mathrm{~T}$
Explanation:
B Given that $\mathrm{i}_{1}=\mathrm{i}_{2}=\mathrm{i} \quad \mathrm{B}_{1}=0.6 \mathrm{~T}$ $r_{1}=r, \quad r_{2}=3 r$ We know that magnetic field at a distance $r$ for a long wire carrying current $B=\frac{\mu_{0} i}{2 \pi r}$ $B \propto \frac{1}{r}$ Now ratio of magnetic field at distance $r_{1}$ and $r_{2}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}} =\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ $\frac{0.6}{\mathrm{~B}_{2}} =\left(\frac{3 \mathrm{r}}{\mathrm{r}}\right)$ $\mathrm{B}_{2} =\frac{0.6}{3}$ $\mathrm{~B}_{2} =0.2 \mathrm{~T}$
J and K CET- 1998
Moving Charges & Magnetism
153318
A long straight wire carries a current $10 \mathrm{~A}$ and has a length of $0.2 \mathrm{~m}$. The wire is positioned perpendicular to a magnetic field. If the wire experiences a force of $0.1 \mathrm{~N}$, the strength of the magnetic field is
1 $1.5 \mathrm{~T}$
2 $0.05 \mathrm{~T}$
3 $0.5 \mathrm{~T}$
4 zero
Explanation:
B Given that $\mathrm{i}=10 \mathrm{~A} \quad l=0.2 \mathrm{~m} \quad \theta=90^{\circ}$ $\mathrm{F}=0.1 \mathrm{~N}$ We know that $\mathrm{F}=\mathrm{iB} l \sin \theta$ $0.1=10 \times \mathrm{B} \times 0.2 \sin 90^{\circ}$ $0.1=10 \times \mathrm{B} \times 0.2 \times 1$ $\mathrm{~B}=0.05 \mathrm{~T}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Moving Charges & Magnetism
153314
A one metre long wire is lying at right angles to the magnetic field. A force of $1 \mathrm{~kg} \mathrm{wt}$. is acting on it in a magnetic field of 0.98 tesla. The current flowing in it will be
1 $100 \mathrm{~A}$
2 $10 \mathrm{~A}$
3 $1 \mathrm{~A}$
4 zero
Explanation:
B Given that $\mathrm{B}=0.98 \mathrm{~T}$ $l=1 \mathrm{~m}$ $\mathrm{~F}=1 \mathrm{~kg} \mathrm{wt}=1 \times 9.8 \mathrm{~N}$ $\theta=90^{\circ}$ We know that, Magnetic force $(F)=i B l \operatorname{Sin} \theta$ $\mathrm{F}=\mathrm{i} \mathrm{B} l \sin 90^{\circ}$ $\mathrm{F}=\mathrm{iB} l$ $1 \times 9.8=\mathrm{i} \times 0.98 \times 1$ $\mathrm{i}=10 \mathrm{amp} .$
J and K CET- 2004
Moving Charges & Magnetism
153316
A straight thin conductor is bent as shown in the adjoining figure. It carries a current $I$ ampere. The radius of the semicircular arc is $r$ meter. The magnetic induction at the centre of semicircular arc is
1 $\frac{\mu_{0}}{4 \pi} \frac{I}{r}$ tesla
2 $\frac{\mu_{0}}{4} \frac{I}{r}$ tesla
3 $\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}}{2 \mathrm{r}}$ tesla
4 zero
Explanation:
B As we know that, Magnetic field for a circular wire $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \mathrm{r}}$ For a semi circle Magnetic field $\left(\mathrm{B}^{\prime}\right)=\mathrm{B} / 2$ $\mathrm{B}^{\prime}=\frac{\mu_{0} \mathrm{I}}{4 \mathrm{r}} \text { Tesla }$
J and K CET- 2001
Moving Charges & Magnetism
153317
The magnetic field at a distance $r$ from a long wire carrying current $i$ is 0.6 Tesla. The magnetic field at distance $3 r$ is
1 $0.1 \mathrm{~T}$
2 $0.2 \mathrm{~T}$
3 $0.9 \mathrm{~T}$
4 $1.8 \mathrm{~T}$
Explanation:
B Given that $\mathrm{i}_{1}=\mathrm{i}_{2}=\mathrm{i} \quad \mathrm{B}_{1}=0.6 \mathrm{~T}$ $r_{1}=r, \quad r_{2}=3 r$ We know that magnetic field at a distance $r$ for a long wire carrying current $B=\frac{\mu_{0} i}{2 \pi r}$ $B \propto \frac{1}{r}$ Now ratio of magnetic field at distance $r_{1}$ and $r_{2}$ $\frac{\mathrm{B}_{1}}{\mathrm{~B}_{2}} =\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ $\frac{0.6}{\mathrm{~B}_{2}} =\left(\frac{3 \mathrm{r}}{\mathrm{r}}\right)$ $\mathrm{B}_{2} =\frac{0.6}{3}$ $\mathrm{~B}_{2} =0.2 \mathrm{~T}$
J and K CET- 1998
Moving Charges & Magnetism
153318
A long straight wire carries a current $10 \mathrm{~A}$ and has a length of $0.2 \mathrm{~m}$. The wire is positioned perpendicular to a magnetic field. If the wire experiences a force of $0.1 \mathrm{~N}$, the strength of the magnetic field is
1 $1.5 \mathrm{~T}$
2 $0.05 \mathrm{~T}$
3 $0.5 \mathrm{~T}$
4 zero
Explanation:
B Given that $\mathrm{i}=10 \mathrm{~A} \quad l=0.2 \mathrm{~m} \quad \theta=90^{\circ}$ $\mathrm{F}=0.1 \mathrm{~N}$ We know that $\mathrm{F}=\mathrm{iB} l \sin \theta$ $0.1=10 \times \mathrm{B} \times 0.2 \sin 90^{\circ}$ $0.1=10 \times \mathrm{B} \times 0.2 \times 1$ $\mathrm{~B}=0.05 \mathrm{~T}$
