What is the dimensions of magnetic field B in terms of C -coulomb- M- L- T-displaystyle left- - M - 1 - L - 1 - T - -2 -C right- displaystyle left- - M - 1 - L - 0 - T - -1 - C - -1 - right- displaystyle left- - M - 1 - L - 0 - T - -2 -C right- displaystyle left- - M - 1 - L - 0 - T - -1 -C right-

Question

Toppr Admin · Accepted Answer

We know that force on a particle moving perpendicular to magnetic field is
$F = q v B$
$\Rightarrow B = \frac{F}{q v}$
Dimension of $F$ is $[M^{1} L^{1} T^{- 2}]$
Dimension of $q$ is $[C^{1}]$
Dimension of $v$ is $[L^{1} T^{- 1}]$
Hence, the dimensions of $B$ is $[M^{1} L^{0} T^{- 1} C^{- 1}]$

List - I	List - II
A) Spring constant	I) [ $M^{1} L^{2} T^{- 2}$ ]
B) Pascal	II) [ $M^{0} L^{0} T^{- 1}$ ]
C) Hertz	III)[ $M^{1} L^{0} T^{- 2}$ ]
D) Joule	IV) [ $M^{1} L^{- 1} T^{- 2}$ ]

List-I	List-II
A.Spring constant	1. $M^{1} L^{2} T^{- 2}$
B.pascal	2. $M^{0} L^{0} T^{- 1}$
C.hertz	3. $M^{1} L^{0} T^{- 2}$
D.joule	4. $M^{1} L^{- 1} T^{- 2}$

What is the dimensions of magnetic field B in terms of C (=coulomb), M, L, T?[M1L1T−2C][M1L0T−1C−1][M1L0T−2C][M1L0T−1C]

We know that force on a particle moving perpendicular to magnetic field is F=qvB⇒B=FqvDimension of F is [M1L1T−2]Dimension of q is [C1]Dimension of v is [L1T−1]Hence, the dimensions of B is [M1L0T−1C−1]

What is the dimensions of magnetic field B in terms of C (=coulomb), M, L, T?
$[M^{1} L^{1} T^{- 2} C]$
$[M^{1} L^{0} T^{- 1} C^{- 1}]$
$[M^{1} L^{0} T^{- 2} C]$
$[M^{1} L^{0} T^{- 1} C]$

We know that force on a particle moving perpendicular to magnetic field is
$F = q v B$
$\Rightarrow B = \frac{F}{q v}$
Dimension of $F$ is $[M^{1} L^{1} T^{- 2}]$
Dimension of $q$ is $[C^{1}]$
Dimension of $v$ is $[L^{1} T^{- 1}]$
Hence, the dimensions of $B$ is $[M^{1} L^{0} T^{- 1} C^{- 1}]$